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Chapter 2

Quantitative Ability

TU CMAT entrance past exam questions

Past questions and answers

214 questions set from this chapter, 17 of them more than once; 1 is most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 11 exams
  • Asked 3 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark
  • 2015 · CMAT · 1 mark

A salary of an officer being increased by 10% every year becomes Rs. 48400 in the third year. What was his original salary? (a) Rs. 45,000 (b) Rs. 44,000 (c) Rs. 42,000 (d) Rs. 40,000

Answer

(d) Rs. 40,000

Let the original salary be Rs. PP.
After two yearly increases of 10%, the salary in the third year is:

P(1+10100)2=48400P\left(1+\frac{10}{100}\right)^2 = 48400

P(1.1)2=48400P(1.1)^2 = 48400

P×1.21=48400P \times 1.21 = 48400

P=484001.21=40000P = \frac{48400}{1.21} = 40000

So, the original salary was Rs. 40,000.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • CMAT Set 02 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 09 · 1 mark
  • 2018 · CMAT (BBM) · 1 mark
  • 2075 · CMAT · 1 mark

If 12 persons can complete a work in 4 days, in how many days can 8 persons complete that work? (a) 4 (b) 6 (c) 8 (d) 10

Answer

(b) 6

Total work =12×4=48= 12 \times 4 = 48 person-days. With 8 persons, days =488=6= \dfrac{48}{8} = 6 days.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • CMAT Set 03 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2018 · CMAT (BBM) · 1 mark
  • 2075 · CMAT · 1 mark

A train 300 m long is moving at a speed of 54 km/hr and it took 40 seconds to cross a bridge. The length of the bridge is (a) 250 m (b) 300 m (c) 325 m (d) 350 m

Answer

(b) 300 m

Speed = 54×518=1554 \times \frac{5}{18} = 15 m/s.
Total distance covered in 40 s = 15×40=60015 \times 40 = 600 m.
Bridge length = 600−300=300600 - 300 = 300 m.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • CMAT Set 03 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

If a sum of money is divided equally among n children, each child will receive Rs. 60. If another child is added to the group, then when the sum is divided equally among all children, each child will receive Rs 50. What is the sum of money? (a) Rs. 100 (b) Rs. 120 (c) Rs. 300 (d) Rs. 400

Answer

(c) Rs. 300

Let the total sum be SS and the number of children be nn.

S=60nandS=50(n+1)S = 60n \quad \text{and} \quad S = 50(n+1)

60n=50n+50⇒10n=50⇒n=560n = 50n + 50 \Rightarrow 10n = 50 \Rightarrow n = 5

S=60×5=Rs. 300S = 60 \times 5 = \text{Rs. } 300

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • CMAT Set 04 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 08 · 1 mark
  • 2018 · CMAT (BBM) · 1 mark
  • 2075 · CMAT · 1 mark

A card is drawn at random from the pack of 52 cards. What is the probability of getting a face card? (a) 4/13 (b) 1/4 (c) 1/2 (d) 3/13

Answer

(d) 3/13

There are 12 face cards (Jack, Queen, King in 4 suits) in a pack of 52 cards.
So, probability = 1252=313\frac{12}{52} = \frac{3}{13}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2018 · CMAT (BBM) · 1 mark
  • 2075 · CMAT · 1 mark

The factors of x4+4x^4 + 4 are (a) (x2+2)(x2−2)(x^2 + 2)(x^2 - 2) (b) (x2+2x+2)(x2−2x+2)(x^2 + 2x + 2)(x^2 - 2x + 2) (c) (x2+2x+2)(x2−2x+2)(x^2 + 2x + 2)(x^2 - 2x + 2) (d) None of these

Answer

(b) (x2+2x+2)(x2−2x+2)(x^2 + 2x + 2)(x^2 - 2x + 2)

Using Sophie Germain’s identity: x4+4=(x2+2)2−(2x)2=(x2+2x+2)(x2−2x+2)x^4+4=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

A sum lent out at 12% per annum compound interest for two years amounts to Rs. 15660. The sum lent is (a) Rs. 12,000 (b) Rs. 12,250 (c) Rs. 12,484 (d) Rs. 12,880

Answer

(c) Rs. 12,484

Using A=P(1+r)nA=P(1+r)^n: 15660=P(1+0.12)2=P(1.12)2=P(1.2544)15660=P(1+0.12)^2=P(1.12)^2=P(1.2544) P=156601.2544=12484 (approximately)P=\frac{15660}{1.2544}=12484 \text{ (approximately)}

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

The interest rate that doubles a sum in 10 years is (a) 12% (b) 11% (c) 10% (d) 9%

Answer

(c) 10%

For simple interest, the amount doubles when total interest equals the principal:

P×R×T=P⇒R×10=100⇒R=10%P \times R \times T = P \Rightarrow R \times 10 = 100 \Rightarrow R = 10\%

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

Ramesh invested a sum of money at simple interest at a certain rate of interest for three years. Had it been invested at a 4% higher rate, it would have fetched Rs. 480 more. Find out the principal amount that was invested by Ramesh? (a) Rs. 3000 (b) Rs. 4000 (c) Rs. 5000 (d) Rs. 4500

Answer

(b) Rs. 4000

Extra interest in 3 years at 4% higher rate: P×4100×3=480P \times \frac{4}{100} \times 3 = 480 P=480×10012=4000P = \frac{480 \times 100}{12} = 4000

So, the principal amount was Rs. 4000.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

When Usha was thrice as old as Nisha her sister Asha was 25. When Nisha was half as old as Asha, then sister Usha was 34. Their ages add to 100. How old is Usha? (a) 20 (b) 30 (c) 40 (d) 50

Answer

(c) 40

Let current ages be Usha = UU, Nisha = NN, Asha = AA.

When Usha was thrice Nisha’s age, Asha was 25: U+x=3(N+x),A+x=25U+x=3(N+x),\quad A+x=25 This gives: U+2A−3N=50(1)U+2A-3N=50 \quad (1)

When Nisha was half Asha’s age, Usha was 34: 2N+2y=A+y,U+y=342N+2y=A+y,\quad U+y=34 This gives: U+A−2N=34(2)U+A-2N=34 \quad (2)

Also, U+N+A=100(3)U+N+A=100 \quad (3)

From (2), (U+A=34+2N). Using (3): 34+2N+N=100⇒3N=66⇒N=2234+2N+N=100 \Rightarrow 3N=66 \Rightarrow N=22

Then (U+A=78). From (1): U+2A−3(22)=50⇒U+2A=116U+2A-3(22)=50 \Rightarrow U+2A=116

Subtracting (U+A=78): A=38,U=40A=38,\quad U=40

So Usha is 40 years old.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

The average of a family of five member is 24 years. If the present age of the youngest member is 8 years, what was the average age of the family at the time of the birth of the youngest member? (a) 50 (b) 40 (c) 30 (d) 20

Answer

(d) 20

Total age of the family now =5×24=120= 5 \times 24 = 120 years.
The youngest member is 8 years old now, so 8 years ago the other four members were each 8 years younger.

Total age of the other four at that time: 120−8−(4×8)=120−8−32=80120 - 8 - (4 \times 8) = 120 - 8 - 32 = 80

At the time of the youngest member’s birth, the family (excluding the newborn) had 4 members, so average age: 804=20\frac{80}{4} = 20

So the average age was 20 years.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

The average marks of a class in a science test is 75. Geeta was the topper with 96 marks. If her mark was wrongly entered as 69, the average of the class decreases by 1. Find the number of students in the class? (a) 24 (b) 25 (c) 27 (d) 29

Answer

(c) 27

Decrease in total marks = 96−69=2796 - 69 = 27.
If the average decreases by 1, then:

27n=1⇒n=27\frac{27}{n} = 1 \Rightarrow n = 27

So, the number of students is 27.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

In an examination, a student's average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination? (a) 12 (b) 13 (c) 11 (d) 14

Answer

(c) 11

Total extra marks = 20 + 2 = 22.
Increase in average = 65 − 63 = 2.
Number of subjects = 22 ÷ 2 = 11.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then (a) 2x=y+z (b) x=y+z (c) x=2y+2z (d) None of these

Answer

(a) 2x=y+z2x=y+z

Total of six numbers =6x=6x. Total of first three =3y=3y and of remaining three =3z=3z.
So 6x=3y+3z⇒2x=y+z6x=3y+3z \Rightarrow 2x=y+z.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, What was the number of valid votes that the other candidate got? (a) 2800 (b) 2700 (c) 2100 (d) 2500

Answer

(b) 2700

Valid votes = 80% of 7500 = 6000.
The other candidate got 45% of valid votes = 0.45×6000=27000.45 \times 6000 = 2700.

Answer checked by Claude Haiku · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

The population of a town increased from 175000 to 2652500 in a decade. What is the average percent increase of population per year? (a) 4% (b) 6% (c) 5% (d) 50%

Answer

(c) 5%

The printed figure 2652500 looks like a typing error for 265,250 (a rise of about 51.6% in ten years is the only reading that fits the options).

Total increase=265,250−175,000=90,250\text{Total increase}=265{,}250-175{,}000=90{,}250

Percent increase=90,250175,000×100≈51.6%\text{Percent increase}=\frac{90{,}250}{175{,}000}\times 100\approx 51.6\%

Average per year=51.6%10≈5.16%≈5%\text{Average per year}=\frac{51.6\%}{10}\approx 5.16\%\approx 5\%

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Asked 2 times
  • 2074 · CMAT · 1 mark
  • 2016 · CMAT · 1 mark

A vendor bought bananas at 6 for a ruppe how many for a rupee must he sell to gain 20%? (a) 3 (b) 4 (c) 5 (d) 6

Answer

(c) 5

Cost price of 1 banana = Rs 16\frac{1}{6}.
To gain 20%, selling price of 1 banana = 16×1.20=15\frac{1}{6} \times 1.20 = \frac{1}{5} rupee.
So, in Rs 1 he must sell 5 bananas.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark
  • 2016 · CMAT · 1 mark

Arun obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark? (a) 53 (b) 54 (c) 72 (d) 75

Similar questions: Arun obtaind 76, 65, 82, 67 and (2074 · CMAT)

Answer

(d) 75

Average =76+65+82+67+855=3755=75=\frac{76+65+82+67+85}{5}=\frac{375}{5}=75.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2074 · CMAT · 1 mark

Arun obtaind 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is average mark? (a) 53 (b) 54 (c) 72 (d) 75

Similar questions: Arun obtained 76, 65, 82, 67 and (2016 · CMAT)

Answer

(d) 75

Average =76+65+82+67+855=3755=75=\dfrac{76+65+82+67+85}{5}=\dfrac{375}{5}=75 marks.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

The difference in simple and compound interest in a certain sum in 2 years at 15% per annum is Rs. 144. The sum is (a) Rs. 6000 (b) Rs. 6200 (c) Rs. 6400 (d) Rs. 6500

Similar questions: The difference between simple and compound interest (2014 · CMAT) · The simple interest on a certain sum (2018 · CMAT (BBM))

Answer

(c) Rs. 6400

For 2 years,
[ \text{CI} - \text{SI} = P\left(\frac{r}{100}\right)^2 ] [ 144 = P\left(\frac{15}{100}\right)^2 = P(0.15)^2 ] [ P = \frac{144}{0.0225} = 6400 ]

So the sum is Rs. 6400.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 09 · 1 mark
  • 2075 · CMAT · 1 mark

A merchant earns 20% by selling an article for Rs. 480. What % would he have gained or lost if he had sold it for Rs. 430? (a) 7.5% Gain (b) 7.5% Loss (c) 6.5% Gain (d) 6.5% Loss

Answer

(a) 7.5% Gain

Cost price =4801.20=Rs. 400= \dfrac{480}{1.20} = \text{Rs. }400

If sold for Rs. 430, gain =430−400=Rs. 30= 430 - 400 = \text{Rs. }30

Gain % =30400×100=7.5%= \dfrac{30}{400} \times 100 = 7.5\% gain.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2075 · CMAT · 1 mark

The area of circle is given by: (a) 2πr2\pi r (b) πr2\pi r^2 (c) πr2h\pi r^2h (d) 2πr22\pi r^2

Answer

(b) πr2\pi r^2

The area of a circle is πr2\pi r^2, where rr is the radius. Note that 2πr2\pi r is the circumference (perimeter), while πr2h\pi r^2 h is the volume of a cylinder.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2015 · CMAT · 1 mark

A grocer has a sale of Rs 6435, Rs 6927, Rs 6855, Rs 7230 and Rs 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs 6500? (a) 4800 (b) 4991 (c) 5001 (d) 5000

Answer

(b) 4991

Total required for 6 months =6500×6=39000= 6500 \times 6 = 39000

Total for first 5 months =6435+6927+6855+7230+6562=34009= 6435 + 6927 + 6855 + 7230 + 6562 = 34009

Sixth month sale =39000−34009=4991= 39000 - 34009 = 4991

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2014 · CMAT · 1 mark

The slope of the line determined by the equation 3x+y=43x + y = 4 is (a) -3 (b) 1/31/3 (c) 3/43/4 (d) 3

Answer

(a) -3

Rewrite the equation as y=−3x+4y = -3x + 4. Comparing with y=mx+cy = mx + c, the slope is m=−3m = -3.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark
  • 2014 · CMAT · 1 mark

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green? (a) 1/3 (b) 3/4 (c) 7/9 (d) 8/21

Answer

(a) 1/3

Total balls = 8 + 7 + 6 = 21.
Neither red nor green means the ball must be blue.
Favourable outcomes = 7 blue balls.
So, probability = 7/21 = 1/3.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2014 · CMAT · 1 mark

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is (a) Rs. 650 (b) Rs. 690 (c) Rs. 698 (d) Rs. 700

Similar questions: A sum of money amounts to Rs.

Answer

(c) Rs. 698

Interest for 1 year = Rs. 854 − Rs. 815 = Rs. 39.
So principal = Rs. 815 − 3 × Rs. 39 = Rs. 815 − Rs. 117 = Rs. 698.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 08 · 1 mark
  • 2014 · CMAT · 1 mark

A school committee consists of 2 teachers and 4 students. The number of different committees that can be formed from 5 teachers and 10 students is (a) 8 (b) 10 (c) 15 (d) 2100

Answer

(d) 2100

Number of committees =(52)×(104)=10×210=2100= \binom{5}{2}\times\binom{10}{4}=10\times210=2100.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2018 · CMAT (BBM) · 1 mark

Average age of a class of 26 students is 14 years. A new student joined the class and the average age of the class increased by 4 months. The age of the new student is (a) 23 years (b) 21 years (c) 19 years (d) 18 years

Answer

(a) 23 years

Old total age = 26×14=36426 \times 14 = 364 years.
Increase in average = 4 months = 13\frac{1}{3} year.
New average = 14+13=43314 + \frac{1}{3} = \frac{43}{3} years.
New total age of 27 students = 27×433=38727 \times \frac{43}{3} = 387 years.
Age of new student = 387−364=23387 - 364 = 23 years.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 08 · 1 mark
  • 2075 · CMAT · 1 mark

The ratio of chicken to pigs to horses in a farm is 10:2:3. If there are 120 chickens in the farm then the number of horses in the farm is (a) 25 (b) 36 (c) 40 (d) 24

Answer

(b) 36

Chickens = 10 parts = 120, so 1 part = 12.
Horses = 3 parts = 3×12=363 \times 12 = 36.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • 2075 · CMAT · 1 mark

The slope of the straight line 4x+3y=104x + 3y = 10 is (a) 4 (b) 3 (c) -4/3 (d) -3/4

Answer

(c) −43-\frac{4}{3}

Rearranging 4x+3y=104x+3y=10 gives 3y=−4x+103y=-4x+10, so y=−43x+103y=-\frac{4}{3}x+\frac{10}{3}. Hence the slope is −43-\frac{4}{3}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2075 · CMAT · 1 mark

If the product of two integers is odd, the sum of those two integers must be (a) odd (b) even (c) prime (d) a perfect square

Answer

(b) even

If the product of two integers is odd, both integers must be odd. The sum of two odd integers is always even.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2014 · CMAT · 1 mark

A homeowner is planning to use carpet tiles to cover the floor of a room measuring 9 feet by 10 feet 8 inches. If the carpet tiles are 8 inches wide and 1 foot long and there are no gaps between the tiles as they are placed on the floor, how many carpet tiles will the homeowner need to cover the floor of the room? (a) 100 (b) 135 (c) 144 (d) 150

Answer

(c) 144

Room area: (9 \text{ ft} \times 10\text{ ft }8\text{ in} = 9 \times \frac{32}{3} = 96 \text{ sq ft})

Tile area: (8\text{ in} \times 1\text{ ft} = \frac{2}{3}\text{ ft} \times 1\text{ ft} = \frac{2}{3}\text{ sq ft})

Number of tiles:
[ \frac{96}{2/3}=96 \times \frac{3}{2}=144 ]

So the homeowner needs 144 carpet tiles.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 08 · 1 mark
  • 2075 · CMAT · 1 mark

A cricket team won 55% of the total number of matches it played during a year. If it lost 30% of the matches played and 3 matches were draw, the total number of matches played by the team during the year was (a) 10 (b) 12 (c) 18 (d) 20

Answer

(d) 20

Won and lost matches together = 55%+30%=85%55\% + 30\% = 85\%, so drawn matches = 100%−85%=15%100\% - 85\% = 15\%.
Given 15%15\% of total matches = 3, so total matches =30.15=20= \frac{3}{0.15} = 20.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2075 · CMAT · 1 mark

If A and B are two sets and A∪B=A∩BA \cup B = A \cap B then (a) A=ϕA = \phi (b) B=ϕB = \phi (c) A∪B=ϕA \cup B = \phi (d) A=BA = B

Answer

(d) A=BA = B

Since A∩B⊆A⊆A∪BA \cap B \subseteq A \subseteq A \cup B, if A∪B=A∩BA \cup B = A \cap B, then all three sets must be equal, so A=BA = B.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 09 · 1 mark
  • 2075 · CMAT · 1 mark

The height of a room is 6 m and the length is twice the breadth. If the area of the floor is 162 m2^2, the area of four walls of the room is (a) 324 m2^2 (b) 360 m2^2 (c) 720 m2^2 (d) 648 m2^2

Answer

(a) 324 m2^2

Let breadth =b= b, so length =2b= 2b.
Floor area: 2b×b=2b2=162⇒b2=81⇒b=92b \times b = 2b^2 = 162 \Rightarrow b^2 = 81 \Rightarrow b = 9 m.
So length =18= 18 m.

Area of four walls =2h(l+b)=2×6×(18+9)=12×27=324= 2h(l+b) = 2 \times 6 \times (18+9) = 12 \times 27 = 324 m2^2.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 08 · 1 mark
  • 2014 · CMAT · 1 mark

The difference between simple and compound interest compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is (a) 625 (b) 630 (c) 640 (d) 650

Similar questions: The difference in simple and compound interest

Answer

(a) 625

For 2 years, the difference between compound interest and simple interest is
D=P(r100)2D = P\left(\frac{r}{100}\right)^2

Given D=1D = 1 and r=4%r = 4\%: 1=P(4100)2=P×16100001 = P\left(\frac{4}{100}\right)^2 = P \times \frac{16}{10000} P=1000016=625P = \frac{10000}{16} = 625

So the sum is Rs 625.

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  • CMAT Set 03 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 08 · 1 mark
  • 2014 · CMAT · 1 mark

Look at this series: 7, 10, 8, 11, 9, 12, .... What number should come next? (a) 7 (b) 10 (c) 12 (d) 13

Answer

(b) 10

The pattern alternates: add 3, subtract 2.

7 + 3 = 10, 10 − 2 = 8, 8 + 3 = 11, 11 − 2 = 9, 9 + 3 = 12, so next 12 − 2 = 10.

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  • CMAT Set 04 · 1 mark
  • CMAT Set 06 · 1 mark
  • 2015 · CMAT · 1 mark

A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday? (a) 290 (b) 304 (c) 285 (d) 270

Answer

(c) 285

A 30-day month starting on Sunday has 5 Sundays and 25 other days.
Total visitors = 5×510+25×240=2550+6000=85505 \times 510 + 25 \times 240 = 2550 + 6000 = 8550.
Average per day = 855030=285\frac{8550}{30} = 285.

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  • CMAT Set 04 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 10 · 1 mark
  • 2075 · CMAT · 1 mark

If y is both the cube and square of an integer and y is between 2 and 100, the value of y is (a) 8 (b) 16 (c) 64 (d) 80

Answer

(c) 64

A number that is both a square and a cube is a sixth power. Between 2 and 100, the only such number is 26=642^6 = 64.

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  • 2020 · CMAT · 1 mark

Which of the following is not a measure of dispersion? (a) Range (b) Standard deviation (c) arithmetic mean (d) variance

Answer

(c) arithmetic mean

Arithmetic mean is a measure of central tendency, not dispersion. Range, standard deviation and variance all measure how spread out the data are.

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  • 2020 · CMAT · 1 mark

The sum of all deviations of the data from its mean is (a) Always positive (b) Always negative (c) zero (d) Non-zero integer

Answer

(c) zero

The sum of the deviations of the observations from their arithmetic mean is always zero, because ∑(x−xˉ)=∑x−nxˉ=∑x−∑x=0\sum (x - \bar{x}) = \sum x - n\bar{x} = \sum x - \sum x = 0.

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  • 2020 · CMAT · 1 mark

Which of the following is incorrect? (a) p(A) > 1 (b) 0 <= p(A) <= 1 (c) p(A)= 0 (d) p(A)+p(A) = 1

Answer

(a) p(A) > 1

The probability of any event AA always satisfies 0≤p(A)≤10 \le p(A) \le 1, so p(A)>1p(A) > 1 is impossible and therefore incorrect. Option (b) states the valid range, and (c) p(A)=0p(A)=0 is valid for an impossible event. Option (d), p(A)+p(A)=1p(A)+p(A)=1, holds only for p(A)=12p(A)=\tfrac12, not in general, but (a) is the statement that can never be true.

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  • 2020 · CMAT · 1 mark

The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is (a) q/p+q+r (b) p/p+q+r (c) p+q/p+q+r (d) p+r/p+q+r

Answer

(b) p/p+q+r

Total marbles = p + q + r, and there are p red marbles, so the probability of selecting a red marble is p/(p + q + r).

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  • 2020 · CMAT · 1 mark

Henry went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Henry winning is 1/9, then the number of tickets bought by Henry is (a) 5 (b) 10 (c) 15 (d) 20

Answer

(c) 15

Probability of winning = tickets bought by Henrytotal tickets sold\dfrac{\text{tickets bought by Henry}}{\text{total tickets sold}}

So, tickets bought by Henry =19×135=15= \dfrac{1}{9} \times 135 = 15.

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  • 2020 · CMAT · 1 mark

The objective of the transportation problem which is to be maximized is to (a) Maximize the total profit (b) Minimize the total loss (c) neither maximize nor minimize (d) Optimal cost

Answer

(a) Maximize the total profit

When the transportation problem is of the maximization type (e.g. profit per unit shipped is given instead of cost), the objective is to find the shipping plan that maximizes the total profit. It is solved by converting it into an equivalent minimization problem — subtracting all profits from the highest profit value to form a "loss" matrix — and then applying the usual transportation method.

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  • 2020 · CMAT · 1 mark

Cramer's rule is also known as ........................ . (a) Inverse matrix method (b) Matrix method (c) determinant method (d) Inverse method

Answer

(c) determinant method

Cramer's rule solves a system of linear equations using determinants of the coefficient matrix and its modified forms, so it is also called the determinant method.

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  • 2020 · CMAT · 1 mark

The lenders are also known as ................ . (a) creditor's (b) debtor's (c) buyer's (d) None of the above

Answer

(a) creditor's

A lender gives money or goods on credit, so the lender is the creditor and the borrower is the debtor. Hence lenders are also known as creditors.

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  • 2020 · CMAT · 1 mark

If any two rows and columns of a determinant are identical, the value of the determinant is ............ . (a) 1 (b) 0 (c) -1 (d) unaltered

Answer

(b) 0

If any two rows or any two columns of a determinant are identical, the determinant becomes zero because interchanging them would change its sign while leaving the value unchanged, which is possible only when the value is 0.

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  • 2020 · CMAT · 1 mark

51 - 46+3(5)= ................ . (a) 40 (b) 20 (c) 41 (d) 52

Answer

(b) 20

51−46+3(5)=51−46+15=5+15=2051 - 46 + 3(5) = 51 - 46 + 15 = 5 + 15 = 20.

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  • 2020 · CMAT · 1 mark

what is the value of the following expression? log(9/14)-log(15/16)+log(35/24) (a) 0 (b) 1 (c) 2 (d) 3

Answer

(a) 0

Using log laws:

log⁡914−log⁡1516+log⁡3524=log⁡(914×1615×3524)\log\frac{9}{14}-\log\frac{15}{16}+\log\frac{35}{24} = \log\left(\frac{9}{14}\times\frac{16}{15}\times\frac{35}{24}\right) =log⁡(915×1624×3514)=log⁡(35×23×52)=log⁡1=0= \log\left(\frac{9}{15}\times\frac{16}{24}\times\frac{35}{14}\right) = \log\left(\frac{3}{5}\times\frac{2}{3}\times\frac{5}{2}\right) = \log 1 = 0

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  • 2020 · CMAT · 1 mark

An association of 2 or more persons to carry on a business for the purpose of earning profit is known as (a) Partnership (b) Legal entity (c) insurance (d) None of above

Answer

(a) Partnership

A partnership is the relation between two or more persons who agree to carry on a business and share its profits. A legal entity is a separate identity in the eyes of law (as a company has), and insurance is a contract of indemnity — neither matches the definition given.

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  • 2020 · CMAT · 1 mark

The factors of 4a2- 16 are (a) 4(a-2)(a+2) (b) 4(a-2)(a+2) (c) (a+2)(a+2) (d) 4(a+2)(a+2)

Answer

(a) 4(a−2)(a+2)

4a2−16=4(a2−4)=4(a−2)(a+2)4a^2-16=4(a^2-4)=4(a-2)(a+2)

using x2−y2=(x−y)(x+y)x^2-y^2=(x-y)(x+y).

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  • 2020 · CMAT · 1 mark

If the maximum value in a series is 25 and its range is 15, the minimum value of the series is. (a) 10 (b) 15 (c) 25 (d) 35

Answer

(a) 10

Range = Maximum − Minimum, so Minimum = 25 − 15 = 10.

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  • 2020 · CMAT · 1 mark

If the square of 1 side of a triangle is equal to the sum of the other 2 sides, then the triangle is (a) Isosceles triangle (b) scalene triangle (c) right angle triangle (d) reflection triangle

Answer

(c) right angle triangle

The statement is the Pythagoras theorem: if the square of one side equals the sum of the squares of the other two sides, the angle opposite that side is 90∘90^\circ, so the triangle is right-angled.

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  • 2020 · CMAT · 1 mark

Unbounded solution in an LPP is ................ . (a) Where the objective function can be decreased indefinitely. (b) Which maximizes the objective function (c) Where the objective function can be increased or decreased indefinitely (d) Where the objective function can be increased indefinitely

Answer

(d) Where the objective function can be increased indefinitely

In a maximisation LPP the solution is unbounded when the feasible region is unbounded in the direction of improvement, so the objective function can be made arbitrarily large without violating any constraint and no finite optimum exists.

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  • 2020 · CMAT · 1 mark

How many permutations of the letter of the word 'EXAM' are there? (a) 600 (b) 120 (c) 246 (d) 24

Answer

(d) 24

The word EXAM has 4 distinct letters, so the number of permutations is 4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24.

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  • 2020 · CMAT · 1 mark

A man positioned at the origin of the coordinate system. The man can take steps of unit measure in the direction North, East, West or South. Find the number of ways of he can reach the point (5,2) covering the shortest possible distance. (a) 252 (b) 432 (c) 452 (d) 21

Answer

(d) 21

To reach ((5,2)) in the shortest possible distance, the man must take exactly:

  • 5 steps East
  • 2 steps North

Total steps (=5+2=7).

Number of ways to arrange these steps:

[ \frac{7!}{5!,2!}=\frac{5040}{120\cdot 2}=21 ]

So the number of ways is 21.

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  • 2020 · CMAT · 1 mark

In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must invest (a) Rs. 3,100 (b) Rs. 6,240 (c) Rs. 6,500 (d) Rs. 9,500

Answer

(b) Rs. 6,240

Income required = Rs. 650.
10% stock gives Rs. 10 income on face value Rs. 100.
So face value needed =65010×100=Rs.6,500= \frac{650}{10} \times 100 = Rs. 6,500.
Market price is Rs. 96 per Rs. 100 face value, so investment =6500×96100=Rs.6,240= 6500 \times \frac{96}{100} = Rs. 6,240.

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  • 2020 · CMAT · 1 mark

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child? (a) 4 years (b) 8 years (c) 10 years (d) None of these

Answer

(a) 4 years

Let the youngest child be xx years old. The ages are x,x+3,x+6,x+9,x+12x, x+3, x+6, x+9, x+12.

5x+30=50⇒x=45x+30=50 \Rightarrow x=4

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  • 2020 · CMAT · 1 mark

An accurate clock shows 8 o'clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon? (a) 144° (b) 150° (c) 168° (d) 180°

Answer

(d) 180°

From 8 o'clock in the morning to 2 o'clock in the afternoon is 6 hours. The hour hand moves 30∘30^\circ per hour, so it rotates 6×30∘=180∘6 \times 30^\circ = 180^\circ.

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  • 2020 · CMAT · 1 mark

How many times are the hour hand and the minute hand of a clock at right angles during their motion from 1:00 pm to 10:00 pm? (a) 9 (b) 10 (c) 18 (d) 20

Answer

(c) 18 (by the usual rule of thumb)

The hands are at right angles twice in every hour, so over the 9 hours from 1:00 pm to 10:00 pm the standard keyed count is 2×9=182\times 9 = 18.

An exact count is slightly different. The hands are perpendicular at t=90+180k5.5t=\dfrac{90+180k}{5.5} minutes after 12:00. For 60≤t≤60060 \le t \le 600, kk runs from 2 to 17, which gives 16 instances (the first at about 1:21:49 pm, the last at about 9:32:44 pm). Since 16 is not among the options, (c) 18 is the intended choice.

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  • 2020 · CMAT · 1 mark

By selling a bicycle for Rs. 2,850, a shopkeeper gains 14%. If the profit is reduced to 8%, then the selling price will be (a) Rs. 2,600 (b) Rs. 2,700 (c) Rs. 2,800 (d) Rs. 3,000

Answer

(b) Rs. 2,700

Cost price = 28501.14=Rs.2,500\frac{2850}{1.14} = Rs. 2,500.
At 8% profit, selling price =2500×1.08=Rs.2,700= 2500 \times 1.08 = Rs. 2,700.

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  • 2018 · CMAT (BBM) · 1 mark

461+461+461+4614^{61} + 4^{61} + 4^{61} + 4^{61} is divisible by (a) 3 (b) 10 (c) 11 (d) 13

Answer

(None of the given options) —
[ 4^{61}+4^{61}+4^{61}+4^{61}=4\cdot4^{61}=4^{62}=2^{124}. ] It is not divisible by (3,10,11,) or (13): [ 4^{62}\equiv 1\pmod 3,\quad 4^{62}\equiv 5\pmod{11},\quad 4^{62}\equiv 3\pmod{13}, ] and it lacks a factor of (5), so it is not divisible by (10). Thus no listed option is correct as written.

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  • 2018 · CMAT (BBM) · 1 mark

125÷25÷5÷2=?125 \div 25 \div 5 \div 2 = ? (a) 12\frac{1}{2} (b) 14\frac{1}{4} (c) 52\frac{5}{2} (d) 25

Answer

(a) 12\frac{1}{2}

Division is done left to right: 125÷25=5125 \div 25 = 5, then 5÷5=15 \div 5 = 1, then 1÷2=121 \div 2 = \frac{1}{2}.

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  • 2018 · CMAT (BBM) · 1 mark

If Suresh secures 50% marks in psychology and 60% marks in sociology, what percent of marks did Suresh get in both subjects together? (a) 50 (b) 55 (c) 60 (d) None of these

Answer

(b) 55

Assuming both subjects carry equal full marks, the combined percentage is the average:

50+602=55%\frac{50+60}{2}=55\%

So Suresh got 55% in both subjects together.

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  • 2018 · CMAT (BBM) · 1 mark

If half of two third of a number added to one thirds of three fourth is x times the number. The value of 'x' is (a) 12/7 (b) 19/12 (c) 7/12 (d) 15/12

Answer

(c) 7/12

Let the number be NN.
Half of two-thirds of NN is 12×23N=N3\frac12 \times \frac23 N = \frac{N}{3}.
One-third of three-fourths of NN is 13×34N=N4\frac13 \times \frac34 N = \frac{N}{4}.
So, N3+N4=4N+3N12=7N12=xN\frac{N}{3} + \frac{N}{4} = \frac{4N+3N}{12} = \frac{7N}{12} = xN.
Hence, x=712x = \frac{7}{12}.

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  • 2018 · CMAT (BBM) · 1 mark

A person bought a horse for Rs. 1440 and sold it for Rs. 1600. What is the profit percent? (a) 11% (b) 10% (c) 12% (d) 15%

Answer

(a) 11%

Profit = Rs. 1600 − Rs. 1440 = Rs. 160

Profit percent = 1601440×100=11.11%≈11%\dfrac{160}{1440} \times 100 = 11.11\% \approx 11\%

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  • 2018 · CMAT (BBM) · 1 mark

A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for Rs. 12 more, he would have made a profit of 20%. Find the cost price of the article. (a) Rs. 80 (b) Rs. 120 (c) Rs. 160 (d) Rs. 180

Answer

None of these: the cost price is Rs 200.

Let the cost price be xx. The first selling price is 0.9x0.9x. The second cost is 0.8x0.8x, and the second selling price is 0.9x+120.9x+12. A 20% profit gives 0.9x+12=1.2(0.8x)=0.96x0.9x+12 = 1.2(0.8x) = 0.96x 0.06x=12⇒x=Rs 2000.06x = 12 \Rightarrow x = \text{Rs } 200 Check: the second purchase costs Rs 160 and sells at Rs 192, and 192/160=1.2192/160 = 1.2, a 20% profit. Rs 200 is not among (a)-(d).

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  • 2018 · CMAT (BBM) · 1 mark

A sum of money amounts to Rs. 5200 in 5 years and to Rs. 5680 in 7 years at simple interest. Find the rate of interest per annum. (a) 4% (b) 5% (c) 6% (d) 2%

Similar questions: The sum of Rs. 550 amounts to

Answer

(c) 6%

Interest for 2 years = Rs. 5680 − Rs. 5200 = Rs. 480
So, interest per year = Rs. 240
Principal = Rs. 5200 − 5 × 240 = Rs. 4000
Rate = 2404000×100=6%\frac{240}{4000}\times 100 = 6\% per annum.

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  • 2018 · CMAT (BBM) · 1 mark

The simple interest on a certain sum for 2 years at 15% per annum is Rs. 144. Find the sum. (a) Rs. 6000 (b) Rs. 6200 (c) Rs. 6400 (d) Rs. 6500

Similar questions: The difference in simple and compound interest

Answer

(None of the above) Rs. 480

[ SI = \frac{P \times R \times T}{100} ]

[ 144 = \frac{P \times 15 \times 2}{100} = \frac{30P}{100} ]

[ P = \frac{144 \times 100}{30} = 480 ]

So the sum is Rs. 480. Since this is not among the given options, the options are inconsistent with the question as stated.
(If the intended question was the difference between CI and SI for 2 years, then the sum would be Rs. 6400, option (c).)

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  • 2018 · CMAT (BBM) · 1 mark

The ratio of the ages of A and B is 3:5. If B is 8 years older to A, then what is A's age? (a) 12 (b) 16 (c) 18 (d) 20

Answer

(a) 12

Let A's age be 3x3x and B's age be 5x5x. Given 5x−3x=8⇒2x=8⇒x=45x - 3x = 8 \Rightarrow 2x = 8 \Rightarrow x = 4. So A's age =3×4=12= 3 \times 4 = 12 years.

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  • 2018 · CMAT (BBM) · 1 mark

A, B and C invest in a business in partnership with B's capital being twice A's and thrice C's. In a year the profit amounts to Rs. 3300 which belongs to the total capital. What is A's capital? (a) 3000 (b) 6000 (c) 9000 (d) 12000

Answer

(c) 9000

Let A’s capital be xx. Then B’s capital =2x=2x, and since B’s capital is thrice C’s, C’s capital =2x3=\frac{2x}{3}.

Total capital =x+2x+2x3=11x3=x+2x+\frac{2x}{3}=\frac{11x}{3}.

Profit Rs 3300 is 10% of total capital, so total capital =3300×10=33000=3300\times 10=33000.

11x3=33000⇒x=9000\frac{11x}{3}=33000 \Rightarrow x=9000

So A’s capital is Rs 9000.

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  • 2018 · CMAT (BBM) · 1 mark

A mixture of 40 liters of milk and water contain 10% water. How much water must be added to make water 20% in the new mixture? (a) 10 ltr (b) 7 ltr (c) 5 ltr (d) 3 ltr

Answer

(c) 5 ltr

Initial water =10%=10\% of 40 =4=4 liters, so milk =36=36 liters.
Let xx liters of water be added.

4+x40+x=20100=15\frac{4+x}{40+x}=\frac{20}{100}=\frac15

5(4+x)=40+x5(4+x)=40+x

20+5x=40+x⇒4x=20⇒x=520+5x=40+x \Rightarrow 4x=20 \Rightarrow x=5

So, 5 liters of water must be added.

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  • 2018 · CMAT (BBM) · 1 mark

A can do a piece of work in 40 days. He worked at it for 5 days, then B finished it in 21 days. In how many days can A and B together finish the work? (a) 10 (b) 12 (c) 15 (d) 18

Answer

(c) 15

A’s 5 days’ work =540=18=\frac{5}{40}=\frac18. Remaining =78=\frac78, done by B in 21 days, so B’s rate =7/821=124=\frac{7/8}{21}=\frac1{24}.

Together: 140+124=3+5120=115\frac1{40}+\frac1{24}=\frac{3+5}{120}=\frac1{15}, so they finish in 15 days.

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  • 2018 · CMAT (BBM) · 1 mark

A car completes a journey in 18 hours. If half of the distance was covered at 40 km/hr and remaining half at 60 km/hr, then the total distance covered was (a) 432 km (b) 450 km (c) 864 km (d) 900 km

Answer

(c) 864 km

Let each half of the distance be dd km. Then d40+d60=18\frac{d}{40}+\frac{d}{60}=18 3d+2d120=18⇒5d120=18⇒d=432\frac{3d+2d}{120}=18 \Rightarrow \frac{5d}{120}=18 \Rightarrow d=432 Total distance =2d=2×432=864=2d=2\times432=864 km.

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  • 2018 · CMAT (BBM) · 1 mark

The difference between the circumference and the diameter of a circle is 210 meters. Find the radius of the circle. (a) 49 meter (b) 48 meter (c) 43 meter (d) None of these

Answer

(a) 49 meter

Circumference C=2πrC = 2\pi r, diameter d=2rd = 2r.

Given: C−d=210C - d = 210 2πr−2r=2102\pi r - 2r = 210 2r(π−1)=2102r(\pi - 1) = 210

Using π=227\pi = \frac{22}{7}: 2r(227−1)=2102r\left(\frac{22}{7} - 1\right) = 210 2r×157=2102r \times \frac{15}{7} = 210 r=210×72×15=49 mr = \frac{210 \times 7}{2 \times 15} = 49 \text{ m}

So the radius is 49 meter.

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  • 2018 · CMAT (BBM) · 1 mark

The radius of a circle is 2 meters. What is the radius of another circle whose area is 9 times that of the first? (a) 216 meter (b) 36 meter (c) 5 meter (d) None of these

Answer

(d) None of these

First circle: r=2r=2 m, so area =πr2=4π=\pi r^2=4\pi m².
Required area =9×4π=36π=9\times 4\pi=36\pi m².
So R2=36⇒R=6R^2=36 \Rightarrow R=6 m. Since 6 m is not among the options, the answer is (d) None of these.

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  • 2018 · CMAT (BBM) · 1 mark

Value of log⁡8÷log⁡88\log_8 \div \log_{\sqrt{8}} 8 is (a) 6log⁡26 \log 2 (b) log⁡2\log 2 (c) 1 (d) -1

Answer

The question is misprinted in the source: the first logarithm has no argument. The usual version is log⁡8\log 8 and log⁡88\log_{\sqrt8}8:

log⁡88=log⁡8log⁡81/2=log⁡812log⁡8=2,log⁡8=3log⁡2\log_{\sqrt8}8=\frac{\log 8}{\log 8^{1/2}}=\frac{\log 8}{\tfrac12\log 8}=2,\qquad \log 8 = 3\log 2

The product log⁡8×log⁡88=3log⁡2×2=6log⁡2\log 8\times\log_{\sqrt8}8 = 3\log2\times2 = 6\log 2, which is option (a); this is most likely the intended question. Taken literally as a division, log⁡8÷log⁡88=32log⁡2\log 8\div\log_{\sqrt8}8=\tfrac32\log2, which matches none of the options.

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  • 2018 · CMAT (BBM) · 1 mark

The value of log⁡216\log_{\sqrt{2}} 16 is (a) 2 (b) 3 (c) 4 (d) 5

Answer

None of the given options — (\log_{\sqrt{2}}16 = \dfrac{\log_2 16}{\log_2 \sqrt{2}} = \dfrac{4}{1/2} = 8).
(If the intended argument was (\sqrt{16}), then the answer would be (c) 4.)

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  • 2018 · CMAT (BBM) · 1 mark

The students of various semester based program in bachelor level under Faculty of Management, TU if one is selected to opt a program, in how many ways a candidate select the program? (a) 11 (b) 7 (c) 1 (d) 21

Answer

(a) 11

Under the Faculty of Management, TU, there are 11 semester-based bachelor-level programs. Since a selected candidate can choose any one of them, the program can be selected in 11 ways.

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  • 2075 · CMAT · 1 mark

The sum of Rs. 5500 amount to Rs. 9460 in 9 years at simple interest. What will be the rate of interest per cent per annum? (a) 7 (b) 8 (c) 9 (d) 6

Answer

(b) 8

Simple interest =9460−5500=3960= 9460 - 5500 = 3960.
Rate =3960×1005500×9=8%= \dfrac{3960 \times 100}{5500 \times 9} = 8\% per annum.

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  • 2075 · CMAT · 1 mark

1+11+11+16=?1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{6}}} = ? (a) 2910\frac{29}{10} (b) 1019\frac{10}{19} (c) 2919\frac{29}{19} (d) 109\frac{10}{9}

Answer

(c) 2919\frac{29}{19}

The intended expression ends in 19\frac{1}{9}: 1+19=1091+\frac19=\frac{10}{9}, then 1+910=19101+\frac{9}{10}=\frac{19}{10}, then 1+1019=29191+\frac{10}{19}=\frac{29}{19}. As printed, with 16\frac16 at the end, the value is 2013\frac{20}{13}, which is not among the options.

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  • 2075 · CMAT · 1 mark

9+99+999+99999 + 99 + 999 + 9999 (a) 11100 (b) 11106 (c) 11007 (d) None of these

Answer

(b) 11106

9+99+999+9999=108+1107+9999=111069+99+999+9999 = 108+1107+9999 = 11106.

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  • 2075 · CMAT · 1 mark

The average of five numbers is 25. If four of the numbers are 30, 29, 25 and 20, the average of the three larger numbers are (a) 26 (b) 27 (c) 28 (d) 29

Answer

(c) 28

Total of five numbers =25×5=125=25\times5=125.
Sum of four given numbers =30+29+25+20=104=30+29+25+20=104.
So the fifth number =125−104=21=125-104=21.

The three larger numbers are 30,29,2530,29,25.
Their average =30+29+253=843=28=\dfrac{30+29+25}{3}=\dfrac{84}{3}=28.

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  • 2075 · CMAT · 1 mark

The average age of a husband, his wife and daughter three years ago was 27 years and that of the wife and daughter five years ago was 20 years. Husband's present age is (a) 35 (b) 40 (c) 32 (d) 45

Answer

(b) 40

Three years ago, total age of husband, wife and daughter = 27×3=8127 \times 3 = 81.
So their present total = 81+3×3=9081 + 3 \times 3 = 90.

Five years ago, total age of wife and daughter = 20×2=4020 \times 2 = 40.
So their present total = 40+2×5=5040 + 2 \times 5 = 50.

Husband’s present age = 90−50=4090 - 50 = 40 years.

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  • 2075 · CMAT · 1 mark

Value of log⁡8+log⁡18\log 8 + \log \frac{1}{8} (a) 6log⁡26 \log 2 (b) log⁡2\log 2 (c) 1 (d) -1

Answer

(None of the above)
[ \log 8+\log\frac18=\log\left(8\cdot\frac18\right)=\log 1=0. ] So the value is (0), not among (a)–(d). Option (a) would be correct only for (\log 8-\log\frac18).

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  • 2075 · CMAT · 1 mark

12, 23, 43, 85, 166, 327, 648 (a) 23 (b) 43 (c) 85 (d) 166

Answer

(b) 43

The pattern is: multiply by 2 and subtract consecutive integers 1, 2, 3, 4, 5, 6.

12×2−1=2312 \times 2 - 1 = 23
23×2−2=4423 \times 2 - 2 = 44 (not 43)
44×2−3=8544 \times 2 - 3 = 85
85×2−4=16685 \times 2 - 4 = 166
166×2−5=327166 \times 2 - 5 = 327
327×2−6=648327 \times 2 - 6 = 648

So the wrong term is 43; it should be 44.

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  • 2075 · CMAT · 1 mark

444, 300, 200, 136, 87, 84, 80 (a) 300 (b) 200 (c) 136 (d) 87

Answer

(d) 87

The differences should be (12^2, 10^2, 8^2, 6^2, 4^2, 2^2):

[ 444-144=300,\quad 300-100=200,\quad 200-64=136 ]

Next should be (136-36=100), not 87. So 87 is the wrong term.

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  • 2075 · CMAT · 1 mark

The value of cot⁡θ\cot \theta when sec⁡θ=4\sec \theta = 4, is (a) 415\frac{4}{\sqrt{15}} (b) 115\frac{1}{\sqrt{15}} (c) −14\frac{-1}{4} (d) 15\sqrt{15}

Answer

(b) 115\frac{1}{\sqrt{15}}

Given sec⁡θ=4\sec\theta=4, so cos⁡θ=14\cos\theta=\frac14.
Then sin⁡θ=1−cos⁡2θ=1−116=154\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\frac1{16}}=\frac{\sqrt{15}}4.
Thus cot⁡θ=cos⁡θsin⁡θ=1/415/4=115\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{1/4}{\sqrt{15}/4}=\frac1{\sqrt{15}}.

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  • 2075 · CMAT · 1 mark

Select the correct statement: (a) Every identity matrix is scalar matrix. (b) Every diagonal matrix is an identity matrix. (c) Every scalar matrix is an identity matrix. (d) A square matrix whose each element is 1 is an identity matrix.

Answer

(a) Every identity matrix is scalar matrix.

An identity matrix has all diagonal elements equal to 1 and all non-diagonal elements 0. A scalar matrix also has all non-diagonal elements 0 and all diagonal elements equal, so the identity matrix is a scalar matrix with scalar value 1.

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  • 2075 · CMAT · 1 mark

There are 7 semester based programs in bachelor level under Faculty of Management, TU. If one can select two programs at a time, in how many ways a candidate can select the program? (a) 14 (b) 7 (c) 1 (d) 21

Answer

(d) 21

Choosing 2 programs out of 7 can be done in
[ \binom{7}{2}=\frac{7\times 6}{2}=21 ] ways.

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  • 2075 · CMAT · 1 mark

Which one is the best measure of dispersion? (a) range (b) mean deviation (c) quartile deviation (d) standard deviation

Answer

(d) standard deviation

Standard deviation is the best measure of dispersion because it uses every observation, is based on the mean, and is mathematically stable and suitable for further statistical analysis. Range, mean deviation and quartile deviation are simpler but less reliable.

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  • 2074 · CMAT · 1 mark

What least number must be added to 1056, so that the sum is completely divisible by 23? (a) 2 (b) 3 (c) 18 (d) 21

Answer

(a) 2

1056÷231056 \div 23 leaves remainder 2121, so adding 23−21=223-21=2 makes it exactly divisible by 23.

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  • 2074 · CMAT · 1 mark

1397×1397=?1397 \times 1397 = ? (a) 1951609 (b) 198709 (c) 1836619 (d) 203719

Answer

(a) 1951609

13972=(1400−3)2=14002−2⋅1400⋅3+32=1960000−8400+9=19516091397^2 = (1400-3)^2 = 1400^2 - 2\cdot1400\cdot3 + 3^2 = 1960000 - 8400 + 9 = 1951609.

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  • 2074 · CMAT · 1 mark

Parash Khadka takes 5 wickets for 26 run in the final match due to which his bowling average imporves by 2. His old bowling average was 14. Find the total number of wickets after the final match. (a) 20 (b) 22 (c) 24 (d) 24

Answer

(b) 22

Old average = 14, so old runs = 14W14W for WW old wickets.
New average improves by 2, so new average = 14−2=1214-2=12.

14W+26W+5=12\frac{14W+26}{W+5}=12 14W+26=12W+6014W+26=12W+60 2W=34⇒W=172W=34 \Rightarrow W=17

Total wickets after the final match =17+5=22=17+5=22.

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  • 2074 · CMAT · 1 mark

A person cross a 600 m long street in 5 minutes. What is his speed in km per hour? (a) 10 (b) 8.4 (c) 3.6 (d) 7.2

Answer

(d) 7.2

Distance = 600 m = 0.6 km, time = 5 minutes = 560=112\frac{5}{60}=\frac{1}{12} hour.
Speed = 0.61/12=0.6×12=7.2\frac{0.6}{1/12}=0.6\times12=7.2 km/h.

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  • 2074 · CMAT · 1 mark

The largest 4 digit number exactly divisible by 88 is: (a) 9944 (b) 9768 (c) 9988 (d) 8888

Answer

(a) 9944

The largest 4-digit number is 9999. Dividing: 9999÷88=1139999 \div 88 = 113 remainder 5555, so the largest multiple of 88 below 9999 is 88×113=994488 \times 113 = 9944.

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  • 2074 · CMAT · 1 mark

(12)3×64+432=?(12)^3 \times 6^4 + 432 = ? (a) 5060 (b) 5184 (c) 5148 (d) 5084

Answer

(b) 5184

As printed, 123×64+432=2,239,92012^3\times 6^4+432 = 2{,}239{,}920, which is not among the options, so the "+" is a typing error for "÷". With division:

123×64432=1728×1296432=1728×3=5184\frac{12^3\times 6^4}{432}=\frac{1728\times 1296}{432}=1728\times 3=5184

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  • 2016 · CMAT · 1 mark

A sum of Rs. 1000 in 2 years on compound interest of 5% per annum. The rate of interest is
(a) 5 (b) 6 (c) 7 (d) 10

Answer

(a) 5

The question omits the final amount. The standard version states that Rs 1000 becomes Rs 1102.50 in 2 years. Using A=P(1+r100)nA=P\left(1+\frac{r}{100}\right)^n:

1102.50=1000(1+r100)2⇒(1+r100)2=1.1025=(1.05)21102.50=1000\left(1+\frac{r}{100}\right)^2 \Rightarrow \left(1+\frac{r}{100}\right)^2=1.1025=(1.05)^2

So 1+r100=1.051+\frac{r}{100}=1.05 and r=5%r=5\% per annum.

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  • 2016 · CMAT · 1 mark

The average of 10 numbers is calculated as 15. It is discovered later on that while calculating the average one number, namely 36, was wrongly read as 26. The correct average is
(a) 14 (b) 14 (c) 18.6 (d) 16

Answer

(d) 16

The wrong sum was (10 \times 15 = 150). Since 36 was read as 26, the sum is 10 less than it should be.
Correct sum (= 150 + 10 = 160), so correct average (= \frac{160}{10} = 16).

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  • 2016 · CMAT · 1 mark

Porky, a fast bowler, takes 5 wickets for 26 runs in the final match due to which his bowling average improves by 2. His old bowling average was 14. Find the total number of wickets after the final match.
(a) 20 (b) 22 (c) 23 (d) 24

Answer

(b) 22

Let his old number of wickets be WW and old runs be RR.
Old average: RW=14⇒R=14W\frac{R}{W}=14 \Rightarrow R=14W.

After taking 5 wickets for 26 runs, new average improves by 2, so it becomes 14−2=1214-2=12:

14W+26W+5=12\frac{14W+26}{W+5}=12

14W+26=12W+60⇒2W=34⇒W=1714W+26=12W+60 \Rightarrow 2W=34 \Rightarrow W=17

Total wickets after the final match =17+5=22=17+5=22.

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  • 2016 · CMAT · 1 mark

What is the 0.5% of 100,000?
(a) 0.05 (b) 0.50 (c) 50 (d) 500

Answer

(d) 500

0.5% of 100,000 = 0.5100×100000=500\frac{0.5}{100} \times 100000 = 500.

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  • 2016 · CMAT · 1 mark

If there are 5 routes for traveling between two places A and B, how many options are there for a person to travel from A to B and return back?
(a) 5 (b) 10 (c) 9 (d) 25

Answer

(d) 25

There are 5 choices for going from A to B and 5 choices for returning from B to A, so total options = 5×5=255 \times 5 = 25.

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  • 2016 · CMAT · 1 mark

If 5a=3b=255a=3b=25, then 30ab=?30ab=?
(a) 100 (b) 150 (c) 625 (d) 1250

Answer

(d) 1250

Since 5a=255a=25, a=5a=5. Also 3b=253b=25, so b=253b=\frac{25}{3}.
Thus ab=5×253=1253ab=5\times \frac{25}{3}=\frac{125}{3}, and
30ab=30×1253=1250.30ab=30\times \frac{125}{3}=1250.

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  • 2015 · CMAT · 1 mark

If the digits of the numbers 5 7 2 6 4 8 9 are arranged in ascending order, how many digits will remain at the same position? (a) None (b) One (c) Two (d) Three

Answer

(d) Three

Ascending order: 2, 4, 5, 6, 7, 8, 9.
Comparing with the original 5, 7, 2, 6, 4, 8, 9, the digits 6, 8 and 9 remain in the same positions.

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  • 2015 · CMAT · 1 mark

If 6 is 24% of a number, which is 40% of the same number? (a) 8 (b) 10 (c) 15 (d) 20

Answer

(b) 10

Let the number be xx. Then 24%24\% of x=6x = 6, so x=60.24=25x = \frac{6}{0.24} = 25.
Now, 40%40\% of 25=0.40×25=1025 = 0.40 \times 25 = 10.

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  • 2015 · CMAT · 1 mark

60 is 80% of which number? (a) 60 (b) 72 (c) 75 (d) 80

Answer

(c) 75

Let the number be xx. Then 80%80\% of x=60x = 60, so 0.8x=60⇒x=600.8=750.8x = 60 \Rightarrow x = \frac{60}{0.8} = 75.

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  • 2015 · CMAT · 1 mark

What 10000 is increased by 0.1% we get, (a) 10001 (b) 10010 (c) 10100 (d) 11000

Answer

(b) 10010

0.1% of 10000 = 0.1100×10000=10\frac{0.1}{100} \times 10000 = 10, so the increased value is 10000+10=1001010000 + 10 = 10010.

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  • 2015 · CMAT · 1 mark

The square root of 130321 is (a) 121 (b) 231 (c) 361 (d) 661

Answer

(c) 361

Since 3612=361×361=130321361^2 = 361 \times 361 = 130321, the square root of 130321 is 361.

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  • 2014 · CMAT · 1 mark

The exam score of 8 students in a class are 68, 72, 75, 80, 85, 90, 92, 95. What is the range of the scores? (a) 27 (b) 25 (c) 23 (d) 20

Answer

(a) 27

Range = highest score − lowest score = 95 − 68 = 27.

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  • 2014 · CMAT · 1 mark

The exam score of 8 students in a class are 68, 72, 75, 80, 85, 90, 92, 95. What is the median score? (a) 80 (b) 82.5 (c) 85 (d) 87.5

Answer

(b) 82.5

The scores are already in ascending order. For 8 values, the median is the average of the 4th and 5th values:
80+852=82.5\frac{80+85}{2}=82.5

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  • 2014 · CMAT · 1 mark

In the first 10 overs of a cricket game, the run rate was 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? (a) 6.25 (b) 6.5 (c) 6.75 (d) 7

Answer

(a) 6.25

Runs in first 10 overs = 10×3.2=3210 \times 3.2 = 32.
Remaining runs needed = 282−32=250282 - 32 = 250 in 40 overs.
Required run rate = 25040=6.25\frac{250}{40} = 6.25.

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  • 2014 · CMAT · 1 mark

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk? (a) 1/40 (b) 1 (c) 40 (d) 80

Answer

(c) 40

40 cows eat 40 bags in 40 days, so 40 cows eat 1 bag per day.
Thus 1 cow eats 140\frac{1}{40} bag per day, and to eat 1 full bag it needs 40 days.

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  • 2014 · CMAT · 1 mark

Simplify: 26÷57=\frac{2}{6} \div \frac{5}{7} = (a) 1235\frac{12}{35} (b) 12\frac{1}{2} (c) 23\frac{2}{3} (d) 35\frac{3}{5}

Answer

(None of the given options) (\frac{2}{6} \div \frac{5}{7} = \frac{1}{3} \times \frac{7}{5} = \frac{7}{15}).

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  • 2014 · CMAT · 1 mark

If 53=31255^3 = 3125, then the value of 5−35^{-3} is (a) 25 (b) 125 (c) 144 (d) 1125\frac{1}{125}

Answer

(d) 1125\frac{1}{125}

A negative exponent means the reciprocal of the power: 5−3=153=11255^{-3} = \dfrac{1}{5^{3}} = \dfrac{1}{125}.

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  • 2014 · CMAT · 1 mark

Seats for Mathematics, Computer and Accounting in a school are in the ratio 5 : 8 : 9. The school has to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats? (a) 2 : 3 : 4 (b) 6 : 7 : 8 (c) 6 : 8 : 9 (d) None of these

Answer

(d) None of these

Increased seats are proportional to:

5×1.40:8×1.50:9×1.75=7:12:15.75=28:48:635 \times 1.40 : 8 \times 1.50 : 9 \times 1.75 = 7 : 12 : 15.75 = 28 : 48 : 63

So the required ratio is 28 : 48 : 63, which is not given in options (a), (b), or (c). Hence, the answer is (d).

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  • 2014 · CMAT · 1 mark

If 6x+5=−x+406x + 5 = -x + 40, then (a) x=−9x = -9 (b) x=−5x = -5 (c) x=5x = 5 (d) x=9x = 9

Answer

(c) x=5x = 5

6x+5=−x+40⇒6x+x=40−5⇒7x=35⇒x=56x + 5 = -x + 40 \Rightarrow 6x + x = 40 - 5 \Rightarrow 7x = 35 \Rightarrow x = 5.

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  • 2014 · CMAT · 1 mark

The three solutions of the equation f(x)=0f(x) = 0 are -2, 0, and 3. Therefore, the three solutions of the equation f(x−2)=0f(x - 2) = 0 are (a) -4, -2, and 1 (b) -2, 0 and 3 (c) 4, 2 and 5 (d) 0, 2 and 5

Answer

(d) 0, 2 and 5

Since the roots of f(x)=0f(x)=0 are −2,0,3-2,0,3, for f(x−2)=0f(x-2)=0 we set
x−2=−2, 0, 3x-2=-2,\ 0,\ 3 So, x=0, 2, 5x=0,\ 2,\ 5

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  • 2014 · CMAT · 1 mark

If the function f is defined by f(x)=x5−1f(x) = x^5 - 1, then f−1f^{-1}, the inverse function of f, is defined by f−1(x)=f^{-1}(x) = (a) 1/x+151/\sqrt[5]{x+1} (b) 1/x+15\sqrt[5]{1/x+1} (c) x+15\sqrt[5]{x+1} (d) x5+1\sqrt[5]{x} + 1

Answer

(c) x+15\sqrt[5]{x+1}

Let y=x5−1y=x^5-1. Then x5=y+1x^5=y+1, so x=y+15x=\sqrt[5]{y+1}. Hence f−1(x)=x+15f^{-1}(x)=\sqrt[5]{x+1}.

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  • 2014 · CMAT · 1 mark

The set of all points (e, t) where t is a real number, is the graph of y=y = (a) 1/et1/e^t (b) e1/te^{1/t} (c) te1/tte^{1/t} (d) ln⁡t\ln t

Answer

(d) ln⁡t\ln t

If the points are (et,t)(e^t,t), then x=etx=e^t and y=ty=t, so t=ln⁡xt=\ln x. Thus the graph is y=ln⁡xy=\ln x (written as y=ln⁡ty=\ln t in the given options).

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  • 2014 · CMAT · 1 mark

Simplify: (Z3)(Z−4)(Z^3)(Z^{-4}) (a) Z3Z^3 (b) Z−1Z^{-1} (c) Z8Z^8 (d) Z−12Z^{-12}

Answer

(b) Z−1Z^{-1}

Using the law of indices am⋅an=am+na^m \cdot a^n = a^{m+n}: Z3⋅Z−4=Z3+(−4)=Z−1Z^3 \cdot Z^{-4} = Z^{3+(-4)} = Z^{-1}

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  • 2014 · CMAT · 1 mark

There are 5 numbers in a data series. First number of the data series is 14 and mean, median and mode are 10, 9 and 6 respectively. Find the 5 numbers: (a) 14, 6, 9, 6, 15 (b) 14, 9, 9, 9, 9 (c) 14, 9, 10, 8, 9 (d) 14, 9, 9, 6, 12

Answer

(a) 14, 6, 9, 6, 15

Sum =14+6+9+6+15=50=14+6+9+6+15=50, so mean =50/5=10=50/5=10.
Arranged in order: 6, 6, 9, 14, 15 → median =9=9.
Mode is 6, since 6 appears twice (more than any other number).

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  • 2014 · CMAT · 1 mark

When a metallic ball bearing is placed inside a cylindrical container, of radius 2 cm, the height of the water, inside the container, increases by 0.6 cm. The radius, to the nearest tenth of a centimeter, of the ball bearing is (a) 1 cm (b) 1.2 cm (c) 2 cm (d) 0.6 cm

Answer

(b) 1.2 cm

Volume of water displaced = volume of ball bearing:

π(2)2(0.6)=43πR3\pi (2)^2(0.6)=\frac{4}{3}\pi R^3

2.4=43R3⇒R3=1.8⇒R≈1.216 cm2.4=\frac{4}{3}R^3 \Rightarrow R^3=1.8 \Rightarrow R\approx 1.216\text{ cm}

To the nearest tenth, the radius is 1.2 cm.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2014 · CMAT · 1 mark

The probability that an electronic device produced by a company does not function properly is equal to 0.1. If 10 devices are bought, then the probability, to the nearest thousandth, that 7 devices function properly is (a) 0.057 (b) 0.470 (c) 0.001 (d) 0

Answer

(a) 0.057

Here, probability that a device functions properly is p=1−0.1=0.9p=1-0.1=0.9.
For exactly 7 out of 10 devices to function properly:

P(X=7)=(107)(0.9)7(0.1)3=120(0.9)7(0.1)3≈0.057P(X=7)=\binom{10}{7}(0.9)^7(0.1)^3=120(0.9)^7(0.1)^3\approx 0.057

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  • 2014 · CMAT · 1 mark

Given a sequence 1, 4, 9, 16, 20, 36, 49, find the odd man out (a) 29 (b) 9 (c) 20 (d) 49

Answer

(c) 20

The sequence is the squares of natural numbers: 12,22,32,42,52,62,72=1,4,9,16,25,36,491^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2 = 1, 4, 9, 16, 25, 36, 49. The term 20 does not fit — it should have been 25. So 20 is the odd man out.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 05 · 1 mark

A sum of money amounts to Rs. 2520 in 2 years and to Rs. 2700 in 5 years at a certain rate of simple interest. The rate of interest is (a) 5% (b) 4% (c) 2.5% (d) 2%

Similar questions: A sum of money at simple interest (2014 · CMAT)

Answer

(c) 2.5%

Interest for 3 years = Rs. 2700 − Rs. 2520 = Rs. 180
So, interest for 1 year = Rs. 180 ÷ 3 = Rs. 60
Principal = Rs. 2520 − (2 × Rs. 60) = Rs. 2400

Rate = 602400×100=2.5%\dfrac{60}{2400} \times 100 = 2.5\%

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  • CMAT Set 01 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

The sum of Rs. 550 amounts to Rs. 946 in 9 years with simple interest. What is the rate of interest percent per annum? (a) 7 (b) 8 (c) 9 (d) 6

Similar questions: A sum of money amounts to Rs. (2018 · CMAT (BBM))

Answer

(b) 8

Interest = Rs. 946 − Rs. 550 = Rs. 396.
Using I=PTR100I=\frac{P T R}{100}:
396=550×9×R100396=\frac{550 \times 9 \times R}{100}
R=396×100550×9=8%R=\frac{396 \times 100}{550 \times 9}=8\% per annum.

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  • CMAT Set 01 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 09 · 1 mark

If the sum of five consecutive odd integers is 735, what is the largest of these integers? (a) 155 (b) 151 (c) 143 (d) 146

Answer

(b) 151

Let the five consecutive odd integers be x−4, x−2, x, x+2, x+4x-4,\ x-2,\ x,\ x+2,\ x+4.
Their sum is 5x=7355x=735, so x=147x=147.
Largest =x+4=147+4=151=x+4=147+4=151.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 10 · 1 mark

What is the greatest prime factor of (24)2−1(2^4)^2 - 1? (a) 3 (b) 5 (c) 11 (d) 17

Answer

(d) 17

(24)2−1=162−1=256−1=255=3×5×17(2^4)^2 - 1 = 16^2 - 1 = 256 - 1 = 255 = 3 \times 5 \times 17, so the greatest prime factor is 17.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 08 · 1 mark

If 6 is 24% of a number, what is 40% of the same number? (a) 8 (b) 10 (c) 15 (d) 20

Answer

(b) 10

Since 24%24\% of the number is 6, the number is 60.24=25\frac{6}{0.24}=25. Then 40%40\% of 25 is 0.40×25=100.40 \times 25 = 10.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 08 · 1 mark

Narasimha, Madhu and Pavan started a business by investing Rs 120,000, Rs 135,000 and Rs 150,000 respectively. Find the share of Pavan out of an annual profit of Rs 56,700. (a) Rs. 16,800 (b) Rs. 18,900 (c) Rs. 21,000 (d) None of these

Answer

(c) Rs. 21,000

Investment ratio =120000:135000:150000=8:9:10=120000:135000:150000=8:9:10.

Pavan's share =108+9+10×56700=1027×56700=21000=\dfrac{10}{8+9+10}\times 56700=\dfrac{10}{27}\times 56700=21000.

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  • CMAT Set 01 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

Given that 100 milliliters is equal to approximately 0.4 cup, 205 milliliters is equal to approximately how many cups? Which of the following expressions models the solution to the problem above? (a) (100−0.4)(205)(100 - 0.4)(205) (b) 105%105\% of 0.40.4 (c) (205−100)(0.4)(205 - 100)(0.4) (d) 205%205\% of 0.40.4

Answer

(d) 205%205\% of 0.40.4

Since 100 mL corresponds to 0.4 cup, 205 mL is 205100=2.05=205%\frac{205}{100}=2.05=205\% of 100 mL, so the number of cups is 205%205\% of 0.4.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 08 · 1 mark

The average of 8 numbers is A, and one of the numbers is 14. If 14 is replaced with 28, then what is the new average in terms of A? (a) A+7/4A + 7/4 (b) A+1/2A + 1/2 (c) A+2A + 2 (d) 2A+12A + 1

Answer

(a) A+74A + \frac{7}{4}

The total of 8 numbers is 8A8A. Replacing 14 with 28 increases the total by 28−14=1428-14=14, so the new average is 8A+148=A+148=A+74\frac{8A+14}{8}=A+\frac{14}{8}=A+\frac{7}{4}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark

The lines 4x+5y+6=04x + 5y + 6 = 0 and 5x−4y+3=05x - 4y + 3 = 0 are (a) Parallel (b) Co-incident (c) intersecting (d) None

Answer

(c) intersecting

The slopes are −45-\frac45 and 54\frac54. Their product is −1-1, so the lines are perpendicular, and hence they intersect each other.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

The cost of carpeting a room at Rs. 50 per square meter is Rs. 1000. Find the breadth of the room if its length is 5 m. (a) 2 m (b) 3 m (c) 2.5 m (d) 4 m

Answer

(d) 4 m

Area of the room =100050=20 m2=\frac{1000}{50}=20\ \text{m}^2.
Since area == length ×\times breadth, breadth =205=4=\frac{20}{5}=4 m.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark

9+99+999+9999=?9 + 99 + 999 + 9999 = ? (a) 11100 (b) 11106 (c) 11007 (d) None of these

Answer

(b) 11106

9+99+999+9999=111069 + 99 + 999 + 9999 = 11106

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 09 · 1 mark

If the cost of electricity increases by 30%, by what percent should one reduce his usage in order that the amount spent on electricity stays the same? (a) 23.07% (b) 25.5% (c) 26% (d) 33.33%

Answer

(a) 23.07%

If the new rate is 1.31.3 times the old rate, usage must become 11.3\frac{1}{1.3} of the old usage.
Reduction =(1−11.3)×100%=0.31.3×100%≈23.07%=\left(1-\frac{1}{1.3}\right)\times100\%=\frac{0.3}{1.3}\times100\%\approx 23.07\%.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark

When 10,000 is increased by 0.1%, we get (a) 10001 (b) 10010 (c) 10100 (d) 11000

Answer

(b) 10010

0.1% of 10,000 = 0.1100×10000=10\frac{0.1}{100} \times 10000 = 10, so the increased value is 10000+10=1001010000 + 10 = 10010.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

Ram obtains 40% of the marks in a paper of 200 marks. Shyam is ahead of Ram by 25% of Ram's marks, while Bhuvan is ahead of Shyam by one ninth of his own marks. How many marks does Bhuvan get? (a) 112.5 (b) 115 (c) 120 (d) 125.25

Answer

(a) 112.5

Ram’s marks = 40%40\% of 200 = 80
Shyam’s marks = 80+25%80 + 25\% of 80 = 80+20=10080 + 20 = 100

Let Bhuvan’s marks be xx.
Bhuvan is ahead of Shyam by x9\frac{x}{9}, so:

x=100+x9x = 100 + \frac{x}{9}

9x=900+x⇒8x=900⇒x=112.59x = 900 + x \Rightarrow 8x = 900 \Rightarrow x = 112.5

So Bhuvan gets 112.5 marks.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark

The total surface area of a cuboid with dimensions l, b and h is (a) 2(lb+bh+lh)2(lb+bh+lh) (b) (lb+bh+lh)(lb+bh+lh) (c) 1/2(lb+bh+lh)1/2(lb+bh+lh) (d) lbhlbh

Answer

(a) 2(lb+bh+lh)2(lb+bh+lh)

A cuboid has six rectangular faces in three pairs — top and bottom each lblb, front and back each bhbh, and the two sides each lhlh. Adding all six faces gives 2(lb+bh+lh)2(lb+bh+lh).

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 10 · 1 mark

18% of the students at a school are in Year 8, and 16% of the students are in Year 9. If the school has 126 Year 8 students, determine the number of Year 9 students. (a) 108 (b) 110 (c) 112 (d) 114

Answer

(c) 112

18% of total = 126, so total students =1260.18=700= \frac{126}{0.18} = 700.
Year 9 students =16%= 16\% of 700=112700 = 112.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 09 · 1 mark

If in a certain school 20 students are taking math and 10 are taking history and 7 are taking both, how many students are there? (a) 30 (b) 23 (c) 37 (d) 27

Answer

(b) 23

Using the inclusion–exclusion principle: n(M∪H)=n(M)+n(H)−n(M∩H)=20+10−7=23n(M \cup H) = n(M) + n(H) - n(M \cap H) = 20 + 10 - 7 = 23.

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  • CMAT Set 01 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 09 · 1 mark

If the lines 2x+3ay−1=02x + 3ay - 1 = 0 and 3x+4y+1=03x + 4y + 1 = 0 are mutually perpendicular, then the value of a is (a) −1/2-1/2 (b) 3 (c) 5 (d) 8

Answer

(a) −1/2-1/2

For perpendicular lines, product of slopes =−1=-1.

Slope of 2x+3ay−1=02x+3ay-1=0 is −23a-\dfrac{2}{3a} and slope of 3x+4y+1=03x+4y+1=0 is −34-\dfrac{3}{4}.

−23a×−34=−1-\frac{2}{3a}\times -\frac{3}{4}=-1 12a=−1⇒a=−12\frac{1}{2a}=-1 \Rightarrow a=-\frac12

So, a=−12a=-\dfrac12.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark

The square root of 130321 is (a) 121 (b) 231 (c) 361 (d) 661

Answer

(c) 361

Since 3612=361×361=130321361^2 = 361 \times 361 = 130321, the square root of 130321 is 361.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 01 · 1 mark

4(xy)3+(x3−y3)2=4(xy)^3 + (x^3 - y^3)^2 = (a) (x3−y3)2(x^3 - y^3)^2 (b) (x3+y3)2(x^3 + y^3)^2 (c) (x3+y3)(x^3 + y^3) (d) (x3−y3)(x^3 - y^3)

Answer

(b) (x3+y3)2(x^3 + y^3)^2

Let a=x3a=x^3 and b=y3b=y^3. Then (xy)3=x3y3=ab(xy)^3=x^3y^3=ab, so 4(xy)3+(x3−y3)2=4ab+(a−b)2=a2+2ab+b2=(a+b)2=(x3+y3)2.4(xy)^3+(x^3-y^3)^2=4ab+(a-b)^2=a^2+2ab+b^2=(a+b)^2=(x^3+y^3)^2.

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  • CMAT Set 01 · 1 mark

If 5a=3b=255a = 3b = 25, then 30ab=?30ab = ? (a) 100 (b) 150 (c) 625 (d) 1250

Answer

(d) 1250

Since (5a=25), (a=5). Also (3b=25), so (b=\frac{25}{3}).
Then (30ab=30\times 5\times \frac{25}{3}=1250).

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark

If 7 workers can assemble a car in 8 hours, how long would it take 12 workers to assemble the same car? (a) 3 (b) 3.5 (c) 4.67 (d) 6.33

Answer

(c) 4.67

Total work = 7×8=567 \times 8 = 56 worker-hours, so 12 workers take 5612=4.67\frac{56}{12} = 4.67 hours.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 08 · 1 mark

The average age of a class of 35 students is 15 years. If the teacher's age is also included, the average age increases by one year. Find the age of the teacher in years. (a) 51 years (b) 54 years (c) 59 years (d) 49 years

Answer

(a) 51 years

Total age of 35 students =35×15=525=35\times15=525.
With teacher, average =16=16 for 36 people, so total =36×16=576=36\times16=576.
Teacher’s age =576−525=51=576-525=51 years.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 09 · 1 mark

By selling an article for Rs. 120, 25% gain is earned. If the same article is sold for Rs. 144, then the gain % will be (a) 25 (b) 40 (c) 45 (d) 50

Answer

(d) 50

Cost price = 1201.25=Rs. 96\frac{120}{1.25} = \text{Rs. }96.
If sold for Rs. 144, gain = 144−96=Rs. 48144-96=\text{Rs. }48, so gain % = 4896×100=50%\frac{48}{96}\times100=50\%.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 10 · 1 mark

The population of a town was 12000 in 1980 and 16000 in 1990. What was the percent increase in the population of the town during this period? (a) 80 (b) 75 (c) 50 (d) 33.3

Answer

(d) 33.3

Increase =16000−12000=4000=16000-12000=4000

Percent increase =400012000×100=33.3%=\dfrac{4000}{12000}\times 100=33.3\%

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 08 · 1 mark

A handkerchief is 20 cm long and 18 cm broad. How much must the breadth be decreased to cover a surface of 324 cm²? (a) 1.8 cm (b) 1.7 cm (c) 1.9 cm (d) 1.6 cm

Answer

(a) 1.8 cm

Original area = 20×18=36020 \times 18 = 360 cm².
Required breadth for area 324 cm² = 32420=16.2\frac{324}{20} = 16.2 cm.
Decrease in breadth = 18−16.2=1.818 - 16.2 = 1.8 cm.

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark

Out of four numbers, the average of the first three is 16 and that of the last three is 15. If the last number is 18, the first number is (a) 20 (b) 21 (c) 23 (d) 25

Answer

(b) 21

First three sum: a+b+c=16×3=48a+b+c=16\times3=48.
Last three sum: b+c+d=15×3=45b+c+d=15\times3=45.
Given d=18d=18, so b+c=45−18=27b+c=45-18=27.
Thus a=48−(b+c)=48−27=21a=48-(b+c)=48-27=21.

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  • CMAT Set 02 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 08 · 1 mark

If 4% of (p+q) is 8 and p is a positive integer, what is the greatest possible value of q? (a) 196 (b) 197 (c) 199 (d) 200

Answer

(c) 199

4% of (p+q)(p+q) is 8:

0.04(p+q)=80.04(p+q)=8

p+q=80.04=200p+q=\frac{8}{0.04}=200

Since pp is a positive integer, its smallest possible value is p=1p=1.
So the greatest possible value of qq is:

q=200−1=199q=200-1=199

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 09 · 1 mark

Rs. 1360 has been divided among A, B, C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. What is B's share? (a) Rs. 120 (b) Rs. 160 (c) Rs. 240 (d) Rs. 320

Answer

(c) Rs. 240

Let B’s share be xx.

Then A gets 23x\frac{2}{3}x, and since B gets 14\frac14 of C, C gets 4x4x.

23x+x+4x=1360\frac{2}{3}x + x + 4x = 1360

173x=1360⇒x=240\frac{17}{3}x = 1360 \Rightarrow x = 240

So B’s share is Rs. 240.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark

If 6x+5=−x+406x + 5 = -x + 40, then x=?x = ? (a) -9 (b) -5 (c) 5 (d) 9

Answer

(c) 5

6x+5=−x+40⇒6x+x=40−5⇒7x=35⇒x=56x + 5 = -x + 40 \Rightarrow 6x + x = 40 - 5 \Rightarrow 7x = 35 \Rightarrow x = 5

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  • CMAT Set 02 · 1 mark

Simplify: (z6)(z−2)(z^6)(z^{-2}) (a) z3z^3 (b) z4z^4 (c) z8z^8 (d) z−12z^{-12}

Answer

(b) z4z^4

Using the law of indices am⋅an=am+na^m \cdot a^n = a^{m+n}: z6⋅z−2=z6+(−2)=z4z^6 \cdot z^{-2} = z^{6+(-2)} = z^4.

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  • CMAT Set 02 · 1 mark
  • CMAT Set 07 · 1 mark

Which of the following numbers is divisible by 12 but not by 8? (a) 72 (b) 88 (c) 108 (d) 120

Answer

(c) 108

108 is divisible by 12 because (108 \div 12 = 9), but it is not divisible by 8 because (108 \div 8 = 13.5).

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 08 · 1 mark

P, Q and R are three consecutive odd numbers in ascending order. If the value of three times P is three less than two times R, find the value of R. (a) 5 (b) 7 (c) 9 (d) 11

Answer

(c) 9

Let the consecutive odd numbers be P=xP=x, Q=x+2Q=x+2, R=x+4R=x+4.

Given: 3P=2R−33P = 2R - 3 3x=2(x+4)−33x = 2(x+4)-3 3x=2x+53x = 2x+5 x=5x=5

So, R=x+4=5+4=9R=x+4=5+4=9.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 06 · 1 mark

On my sister's 15th birthday, she was 159 cm in height, having grown 6% since the year before. How tall was she the previous year? (a) 140 (b) 150 (c) 160 (d) 165

Answer

(b) 150

Let her height the previous year be xx cm. Since she grew 6%,
x×1.06=159x \times 1.06 = 159
x=1591.06=150x = \frac{159}{1.06} = 150
So she was 150 cm tall the previous year.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 03 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 10 · 1 mark

Find the odd one out: 3, 5, 11, 14, 17, 21 (a) 21 (b) 7 (c) 14 (d) 3

Answer

(c) 14
All the given numbers are odd except 14, which is the only even number.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 10 · 1 mark

Suppose p is even and q is odd. Then which of the following cannot be an integer? I. (p+q)/p(p+q)/p II. pq/3pq/3 III. q/p2q/p^2 (a) II only (b) III only (c) I and II only (d) I and III only

Answer

(d) I and III only

  • I. (p+q)/p=1+q/p(p+q)/p = 1 + q/p: Since pp is even and qq is odd, pp cannot divide qq (an even number cannot divide an odd number). So q/pq/p is not an integer, hence (p+q)/p(p+q)/p cannot be an integer.
  • II. pq/3pq/3: This can be an integer. For example, if p=2p=2 and q=3q=3, then pq/3=6/3=2pq/3=6/3=2.
  • III. q/p2q/p^2: Since pp is even, p2p^2 is divisible by 44, so it cannot divide the odd number qq. Hence q/p2q/p^2 cannot be an integer.

Therefore, I and III cannot be integers, so the correct option is (d).

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  • CMAT Set 02 · 1 mark

Value of x in 3x+1+3x=1083^{x+1} + 3^x = 108 is (a) 3 (b) 2 (c) 4 (d) -3

Answer

(a) 3

3x+1+3x=3⋅3x+3x=4⋅3x=1083^{x+1}+3^x=3\cdot3^x+3^x=4\cdot3^x=108, so 3x=27=333^x=27=3^3, giving x=3x=3.

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  • CMAT Set 02 · 1 mark

If 813=2z8^{13} = 2^z, then z=z = (a) 10 (b) 13 (c) 39 (d) 26

Answer

(c) 39

Since 8=238 = 2^3, we have 813=(23)13=2398^{13} = (2^3)^{13} = 2^{39}, so z=39z = 39.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 02 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after the 17th inning. (a) 39 (b) 38 (c) 38.5 (d) 39.5

Answer

(a) 39

Let the average after the 17th inning be xx.
Before the 17th inning, average was x−3x-3, so:

87=17x−16(x−3)=x+4887 = 17x - 16(x-3) = x + 48

x=39x = 39

So his average after the 17th inning is 39 runs.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 09 · 1 mark

Arun spent 25% of his pocket money, and has Rs 125 left. How much had he at first? (a) 120 (b) 144.45 (c) 166.66 (d) 175.67

Answer

(c) 166.66

He spent 25%, so 75% remained. Thus 0.75x=125⇒x=1250.75=166.66…0.75x = 125 \Rightarrow x = \frac{125}{0.75} = 166.66\ldots

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 10 · 1 mark

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is (a) 2:5 (b) 3:5 (c) 4:5 (d) 6:7

Answer

(c) 4:5

Let the third number be 100. Then the first number is 100+20%=120100 + 20\% = 120 and the second is 100+50%=150100 + 50\% = 150.

Ratio =120:150=4:5= 120 : 150 = 4 : 5.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

A train crosses a platform 100 meters long in 60 seconds at a speed of 45 km per hour. The time taken by the train to cross an electric pole is (a) 8 seconds (b) 1 minute (c) 52 seconds (d) 20 seconds

Answer

(c) 52 seconds

Speed =45=45 km/h =45×518=12.5=45\times\frac{5}{18}=12.5 m/s.
In 60 s, train covers 12.5×60=75012.5\times60=750 m.
Platform length =100=100 m, so train length =750−100=650=750-100=650 m.
Time to cross a pole =65012.5=52=\frac{650}{12.5}=52 seconds.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 04 · 1 mark
  • CMAT Set 08 · 1 mark

A jar contains 24 blue balls and 40 red balls. Which one of the following is 50% of the blue balls? (a) 10 (b) 11 (c) 12 (d) 13

Answer

(c) 12

50% of the blue balls means half of 24 blue balls: (24 \div 2 = 12).

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark

125÷25÷5÷2=?125 \div 25 \div 5 \div 2 = ? (a) 12\frac{1}{2} (b) 14\frac{1}{4} (c) 52\frac{5}{2} (d) 25

Answer

(a) 12\frac{1}{2}

Working left to right: 125÷25=5125 \div 25 = 5, then 5÷5=15 \div 5 = 1, then 1÷2=121 \div 2 = \frac{1}{2}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 08 · 1 mark

Two quantities are in the ratio 7:4. If the smaller quantity is 14, then the greater quantity will be: (a) 25 (b) 23 (c) 24.5 (d) None

Answer

(c) 24.5

The smaller quantity 14 corresponds to 4 parts, so 1 part =144=3.5= \frac{14}{4} = 3.5.
Greater quantity =7×3.5=24.5= 7 \times 3.5 = 24.5.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

There are 3 red, 2 blue and 5 white balls in a box. A ball is taken out randomly. What is the probability of a ball being blue or white? (a) 710\frac{7}{10} (b) 310\frac{3}{10} (c) 27\frac{2}{7} (d) 57\frac{5}{7}

Answer

(a) 710\frac{7}{10}

Total balls =3+2+5=10=3+2+5=10. Favorable balls (blue or white) =2+5=7=2+5=7.
So, probability =710=\frac{7}{10}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark

0.3×0.006=0.3 \times 0.006 = (a) 0.00018 (b) 0.0018 (c) 0.018 (d) 0.18

Answer

(b) 0.0018

0.3×0.006=310×61000=1810000=0.00180.3 \times 0.006 = \frac{3}{10} \times \frac{6}{1000} = \frac{18}{10000} = 0.0018

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 08 · 1 mark

The difference between C.I. and S.I. on a sum of Rs. 4800 for 2 years at 5% per annum will be (a) Rs. 10 (b) Rs. 30 (c) Rs. 12 (d) None

Answer

(c) Rs. 12

Difference = P(r100)2=4800×(5100)2=4800×0.0025=Rs. 12P\left(\frac{r}{100}\right)^2 = 4800 \times \left(\frac{5}{100}\right)^2 = 4800 \times 0.0025 = \text{Rs. }12.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark
  • CMAT Set 07 · 1 mark
  • CMAT Set 09 · 1 mark

Let u represent the sum of the integers from 1 through 20, and v represent the sum of the integers from 21 through 40. What is the value of v−uv - u? (a) 39 (b) 200 (c) 320 (d) 400

Answer

(d) 400

u=20×212=210u = \frac{20 \times 21}{2} = 210 and v=40×412−210=820−210=610v = \frac{40 \times 41}{2} - 210 = 820 - 210 = 610.
So, v−u=610−210=400v - u = 610 - 210 = 400.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark

The value of cot⁡θ\cot \theta when sec⁡θ=4\sec \theta = 4 is (a) 154\frac{15}{4} (b) 115\frac{1}{\sqrt{15}} (c) 14\frac{1}{4} (d) 15\sqrt{15}

Answer

(b) 115\frac{1}{\sqrt{15}}

Since sec⁡θ=4\sec\theta=4, we have cos⁡θ=14\cos\theta=\frac14. Then sin⁡θ=1−cos⁡2θ=1−116=154\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\frac1{16}}=\frac{\sqrt{15}}{4} so cot⁡θ=cos⁡θsin⁡θ=1/415/4=115.\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{1/4}{\sqrt{15}/4}=\frac{1}{\sqrt{15}}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark

(12)×(56)×(85)=\left(\frac{1}{2}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{8}{5}\right) = (a) 512\frac{5}{12} (b) 12\frac{1}{2} (c) 23\frac{2}{3} (d) 43\frac{4}{3}

Answer

(c) 23\frac{2}{3}

Cancel the common factor 55: (12)×(56)×(85)=12×16×8=812=23\left(\frac12\right)\times\left(\frac56\right)\times\left(\frac85\right)=\frac{1}{2}\times\frac{1}{6}\times 8=\frac{8}{12}=\frac{2}{3}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark

If 5a=31255^a = 3125, then the value of 5a−35^{a-3} is (a) 25 (b) 125 (c) 625 (d) 1625

Answer

(a) 25

Since 3125=553125 = 5^5, we get a=5a = 5.
So 5a−3=55−3=52=255^{a-3} = 5^{5-3} = 5^2 = 25.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 03 · 1 mark

60 is 80% of which number? (a) 60 (b) 72 (c) 75 (d) 80

Answer

(c) 75

Let the number be xx. Then 80%80\% of x=60x = 60, so 80100x=60⇒x=60×10080=75\frac{80}{100}x=60 \Rightarrow x=\frac{60 \times 100}{80}=75.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark

212+212+212+212=2^{12} + 2^{12} + 2^{12} + 2^{12} = (a) 2142^{14} (b) 4164^{16} (c) 2162^{16} (d) 2482^{48}

Answer

(a) 2142^{14}

212+212+212+212=4⋅212=22⋅212=2142^{12}+2^{12}+2^{12}+2^{12}=4\cdot 2^{12}=2^2\cdot 2^{12}=2^{14}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 09 · 1 mark

If one type of rice costing Rs. 20 per kg and another type of rice costing Rs. 17 per kg are mixed in the ratio 1:2, what is the cost of the mixture per kg? (a) Rs. 17.50 (b) Rs. 18 (c) Rs. 18.50 (d) Rs. 19

Answer

(b) Rs. 18

Using the alligation/weighted average method: [ \text{Cost per kg}=\frac{1\times 20+2\times 17}{1+2} =\frac{20+34}{3} =\frac{54}{3} =Rs.\ 18 ]

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 05 · 1 mark
  • CMAT Set 10 · 1 mark

A mixture of 40 litres of milk and water contains 10% water. How much water must be added to make water 20% in the new mixture? (a) 10 ltr (b) 7 ltr (c) 5 ltr (d) 3 ltr

Answer

(c) 5 ltr

Initial water =10%=10\% of 40 =4=4 litres, so milk =36=36 litres.

Let xx litres of water be added. Then:

4+x40+x=20100=15\frac{4+x}{40+x}=\frac{20}{100}=\frac15 5(4+x)=40+x⇒20+5x=40+x⇒4x=20⇒x=55(4+x)=40+x \Rightarrow 20+5x=40+x \Rightarrow 4x=20 \Rightarrow x=5

So, 5 litres of water must be added.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 05 · 1 mark

John is 20 years older than Steve. In 10 years, Steve's age will be half that of John's. What is Steve's age? (a) 8 (b) 10 (c) 20 (d) 25

Answer

(b) 10

Let Steve's age be SS. John is S+20S+20. In 10 years: S+10=12(S+30)S+10=\frac{1}{2}(S+30) 2S+20=S+30⇒S=102S+20=S+30 \Rightarrow S=10 So Steve is 10 years old.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 06 · 1 mark

What fraction of a week is 98 hours? (a) 7/24 (b) 24/98 (c) 1/2 (d) 7/12

Answer

(d) 7/12

A week has 7×24=1687 \times 24 = 168 hours, so the fraction is 98168=712\frac{98}{168} = \frac{7}{12}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 07 · 1 mark

The average of x and y is 40. If z = 10, what is the average of x, y and z? (a) 16.6 (b) 20 (c) 25 (d) 30

Answer

(d) 30

The average of xx and yy is 40, so x+y=2×40=80x+y=2\times40=80.
Adding z=10z=10 gives 80+10=9080+10=90, and the average of three numbers is 903=30\frac{90}{3}=30.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark

Simplify: (z6)(z−2)(z^6)(z^{-2}) (a) z3z^3 (b) z4z^4 (c) z8z^8 (d) z−12z^{-12}

Answer

(b) z4z^4

Using the law of exponents am⋅an=am+na^m \cdot a^n = a^{m+n}: z6⋅z−2=z6+(−2)=z4z^6 \cdot z^{-2} = z^{6+(-2)} = z^4

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark

9+99+999+9999=?9 + 99 + 999 + 9999 = ? (a) 11100 (b) 11106 (c) 11007 (d) None of these

Answer

(b) 11106

9+99+999+9999=108+999+9999=1107+9999=111069 + 99 + 999 + 9999 = 108 + 999 + 9999 = 1107 + 9999 = 11106

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 08 · 1 mark
  • CMAT Set 09 · 1 mark

The ratio of Rita's age to her mother's age is 3:8. The difference of their ages is 35 years. The ratio of their ages after 4 years will be: (a) 7:12 (b) 5:12 (c) 38:43 (d) 42:47

Answer

(b) 5:12

Rita and her mother are 3x3x and 8x8x. Difference =5x=35=5x=35, so x=7x=7. Their ages are 21 and 56. After 4 years, ratio =25:60=5:12=25:60=5:12.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 04 · 1 mark
  • CMAT Set 06 · 1 mark
  • CMAT Set 10 · 1 mark

If n is an integer, then which one of the following expressions must be even? (a) n2+1n^2+1 (b) n(n+2)n(n+2) (c) n(n+1)n(n+1) (d) n(n+4)n(n+4)

Answer

(c) n(n+1)n(n+1)

Because nn and n+1n+1 are consecutive integers, one of them is always even. Hence their product n(n+1)n(n+1) must always be even.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 05 · 1 mark

0.3 × 0.006 = (a) 0.00018 (b) 0.0018 (c) 0.018 (d) 0.18

Answer

(b) 0.0018

0.3×0.006=310×61000=1810000=0.00180.3 \times 0.006 = \frac{3}{10} \times \frac{6}{1000} = \frac{18}{10000} = 0.0018

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 05 · 1 mark

212+212+212+212=2^{12} + 2^{12} + 2^{12} + 2^{12} = (a) 2142^{14} (b) 4164^{16} (c) 2162^{16} (d) 2482^{48}

Answer

(a) 2142^{14}

212+212+212+212=4⋅212=22⋅212=2142^{12}+2^{12}+2^{12}+2^{12}=4\cdot 2^{12}=2^2\cdot 2^{12}=2^{14}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 05 · 1 mark

60 is 80% of which number? (a) 60 (b) 72 (c) 75 (d) 80

Answer

(c) 75

Let the number be xx. Then 80%80\% of x=60x = 60, so 0.8x=60⇒x=600.8=750.8x = 60 \Rightarrow x = \frac{60}{0.8} = 75.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 05 · 1 mark

4(xy)3+(x3−y3)2=4(xy)^3 + (x^3 - y^3)^2 = (a) (x3−y3)2(x^3 - y^3)^2 (b) (x3+y3)2(x^3 + y^3)^2 (c) (x3+y3)(x^3 + y^3) (d) (x3−y3)(x^3 - y^3)

Answer

(b) (x3+y3)2(x^3 + y^3)^2

Let a=x3a=x^3 and b=y3b=y^3. Then (xy)3=x3y3=ab(xy)^3=x^3y^3=ab, so 4(xy)3+(x3−y3)2=4ab+(a−b)2=a2+2ab+b2=(a+b)2=(x3+y3)2.4(xy)^3+(x^3-y^3)^2=4ab+(a-b)^2=a^2+2ab+b^2=(a+b)^2=(x^3+y^3)^2.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 05 · 1 mark

If 5a=3b=255a = 3b = 25, then 30ab=?30ab = ? (a) 100 (b) 150 (c) 625 (d) 1250

Answer

(d) 1250

Since (5a=25), (a=5). Also (3b=25), so (b=\frac{25}{3}).
Then (30ab=30\times 5\times \frac{25}{3}=1250).

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 05 · 1 mark

If 5a=31255^a = 3125, then the value of 5a−35^{a-3} is (a) 25 (b) 125 (c) 625 (d) 1625

Answer

(a) 25

Since 3125=553125 = 5^5, we get a=5a = 5. So 5a−3=55−3=52=255^{a-3} = 5^{5-3} = 5^2 = 25.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 06 · 1 mark

The value of cot⁡θ\cot \theta when sec⁡θ=4\sec \theta = 4 is (a) 154\frac{15}{4} (b) 115\frac{1}{\sqrt{15}} (c) 14\frac{1}{4} (d) 15\sqrt{15}

Answer

(b) 115\frac{1}{\sqrt{15}}

Since sec⁡θ=4\sec\theta=4, we have cos⁡θ=14\cos\theta=\frac14. Then sin⁡θ=1−cos⁡2θ=154\sin\theta=\sqrt{1-\cos^2\theta}=\frac{\sqrt{15}}4, so cot⁡θ=cos⁡θsin⁡θ=115\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{1}{\sqrt{15}}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 06 · 1 mark

If 6x+5=−x+406x + 5 = -x + 40, then x=?x = ? (a) -9 (b) -5 (c) 5 (d) 9

Answer

(c) 5

6x+5=−x+40⇒6x+x=40−5⇒7x=35⇒x=56x + 5 = -x + 40 \Rightarrow 6x + x = 40 - 5 \Rightarrow 7x = 35 \Rightarrow x = 5

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 06 · 1 mark

When 10,000 is increased by 0.1%, we get (a) 10001 (b) 10010 (c) 10100 (d) 11000

Answer

(b) 10010

0.1% of 10,000 = 0.1100×10000=10\frac{0.1}{100}\times 10000 = 10, so the increased value is 10000+10=1001010000+10=10010.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 07 · 1 mark

Value of x in 3x+1+3x=1083^{x+1} + 3^x = 108 is (a) 3 (b) 2 (c) 4 (d) -3

Answer

(a) 3

3x+1+3x=3x(3+1)=4⋅3x=1083^{x+1}+3^x=3^x(3+1)=4\cdot3^x=108, so 3x=27=333^x=27=3^3, giving x=3x=3.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 07 · 1 mark

If 813=2z8^{13} = 2^z, then z = (a) 10 (b) 13 (c) 39 (d) 26

Answer

(c) 39

Since 8=238 = 2^3, we have 813=(23)13=2398^{13} = (2^3)^{13} = 2^{39}, so z=39z = 39.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 07 · 1 mark

The square root of 130321 is (a) 121 (b) 231 (c) 361 (d) 661

Answer

(c) 361

Since 3612=361×361=130321361^2 = 361 \times 361 = 130321, the square root of 130321 is 361.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 07 · 1 mark

125 ÷ 25 ÷ 5 ÷ 2 = ? (a) 1/2 (b) 1/4 (c) 5/2 (d) 25

Answer

(a) 1/2

Working left to right: (125 \div 25 = 5), then (5 \div 5 = 1), then (1 \div 2 = \frac{1}{2}).

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 07 · 1 mark

(1/2) × (5/6) × (8/5) = (a) 5/12 (b) 1/2 (c) 2/3 (d) 4/3

Answer

(c) 2/3

Cancel the 5s: 12×56×85=12×86=812=23\frac{1}{2}\times\frac{5}{6}\times\frac{8}{5}=\frac{1}{2}\times\frac{8}{6}=\frac{8}{12}=\frac{2}{3}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

4(xy)3+(x3−y3)2=4(xy)^3 + (x^3 - y^3)^2 = a) (x3−y3)2(x^3 - y^3)^2 b) (x3+y3)2(x^3 + y^3)^2 c) (x3+y3)(x^3 + y^3) d) (x3−y3)(x^3 - y^3)

Answer

(b) (x3+y3)2(x^3+y^3)^2

Expand: 4x3y3+(x6−2x3y3+y6)=x6+2x3y3+y6=(x3+y3)24x^3y^3 + (x^6 - 2x^3y^3 + y^6) = x^6 + 2x^3y^3 + y^6 = (x^3+y^3)^2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

60 is 80% of which number? a) 60 b) 72 c) 75 d) 80

Answer

(c) 75

Let the number be (x). Then (80%) of (x = 60), so (\dfrac{80}{100} \times x = 60), giving (x = \dfrac{60 \times 100}{80} = 75).

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

If 813=2z8^{13} = 2^z, then z=z = a) 10 b) 13 c) 39 d) 26

Answer

(c) 39

Since 8=238 = 2^3, we get 813=(23)13=23×13=2398^{13} = (2^3)^{13} = 2^{3 \times 13} = 2^{39}. Comparing with 2z2^z gives z=39z = 39.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

If 5a=31255^a = 3125, then the value of 5(a−3)5^{(a-3)} is a) 25 b) 125 c) 625 d) 1625

Answer

(a) 25

Given 5a=3125=555^a = 3125 = 5^5, so a=5a = 5.

5(a−3)=55−3=52=255^{(a-3)} = 5^{5-3} = 5^2 = 25

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

If 6x+5=−x+406x + 5 = -x + 40, then x=?x = ? a) -9 b) -5 c) 5 d) 9

Answer

(c) 5

6x+5=−x+406x + 5 = -x + 40 6x+x=40−56x + x = 40 - 5 7x=357x = 35 x=5x = 5

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

The square root of 130321 is a) 121 b) 231 c) 361 d) 661

Answer

(c) 361

Check: 3612=130321361^2 = 130321, so 130321=361\sqrt{130321}=361.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

(1/2)×(5/6)×(8/5)=(1/2) \times (5/6) \times (8/5) = a) 5/12 b) 1/2 c) 2/3 d) 4/3

Answer

(c) 2/3

12×56×85=5×82×6×5=812=23\frac12\times\frac56\times\frac85=\frac{5\times 8}{2\times 6\times 5}=\frac{8}{12}=\frac23

Answer checked by Claude Sonnet · Oct 21, 2026

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  • CMAT Set 08 · 1 mark

When 10,000 is increased by 0.1%, we get a) 10001 b) 10010 c) 10100 d) 11000

Answer

(b) 10010

0.1% of 10,000 = 0.1100×10000=10\frac{0.1}{100}\times 10000 = 10, so the increased value is 10000+10=1001010000+10=10010.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • CMAT Set 09 · 1 mark

212+212+212+212=2^{12} + 2^{12} + 2^{12} + 2^{12} = (a) 2142^{14} (b) 4164^{16} (c) 2162^{16} (d) 2482^{48}

Answer

(a) 2142^{14}

212+212+212+212=4⋅212=22⋅212=2142^{12}+2^{12}+2^{12}+2^{12}=4\cdot 2^{12}=2^2\cdot 2^{12}=2^{14}.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 10 · 1 mark

0.3×0.006=0.3 \times 0.006 = (a) 0.00018 (b) 0.0018 (c) 0.018 (d) 0.18

Answer

(b) 0.0018

0.3×0.006=310×61000=1810000=0.00180.3 \times 0.006 = \frac{3}{10} \times \frac{6}{1000} = \frac{18}{10000} = 0.0018.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 10 · 1 mark

Value of x in 3x+1+3x=1083^{x+1} + 3^x = 108 is (a) 3 (b) 2 (c) 4 (d) -3

Answer

(a) 3

3x+1+3x=3x(3+1)=4⋅3x=1083^{x+1}+3^x=3^x(3+1)=4\cdot3^x=108, so 3x=27=333^x=27=3^3, giving x=3x=3.

Answer checked by Claude Haiku · Oct 21, 2026

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  • CMAT Set 10 · 1 mark

125÷25÷5÷2=?125 \div 25 \div 5 \div 2 = ? (a) 12\frac{1}{2} (b) 14\frac{1}{4} (c) 52\frac{5}{2} (d) 25

Answer

(a) 12\frac{1}{2}

Division is done left to right: 125÷25=5125 \div 25 = 5, then 5÷5=15 \div 5 = 1, then 1÷2=121 \div 2 = \frac{1}{2}.

Answer checked by Claude Haiku · Oct 21, 2026

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Questions from CMAT Set 01 (BSc CSIT entrance · CMAT Set 1), CMAT Set 02 (BSc CSIT entrance · CMAT Set 2), CMAT Set 03 (BSc CSIT entrance · CMAT Set 3), CMAT Set 04 (BSc CSIT entrance · CMAT Set 4), CMAT Set 05 (BSc CSIT entrance · CMAT Set 5), CMAT Set 06 (BSc CSIT entrance · CMAT Set 6), CMAT Set 07 (BSc CSIT entrance · CMAT Set 7), CMAT Set 08 (BSc CSIT entrance · CMAT Set 8), CMAT Set 09 (BSc CSIT entrance · CMAT Set 9), CMAT Set 10 (BSc CSIT entrance · CMAT Set 10), CMAT 2020 Past Question (BSc CSIT entrance · CMAT · 2020), CMAT 2019 Past Question (BSc CSIT entrance · CMAT · 2019), CMAT (BBM) 2018 Past Question (BSc CSIT entrance · CMAT (BBM) · 2018), CMAT 2018 Past Question (BSc CSIT entrance · CMAT · 2075), CMAT 2017 Past Question (BSc CSIT entrance · CMAT · 2074), CMAT 2016 Past Question (BSc CSIT entrance · CMAT · 2016), CMAT 2015 Past Question (BSc CSIT entrance · CMAT · 2015), CMAT 2014 Past Question (BSc CSIT entrance · CMAT · 2014), CMAT 2012 Past Question (BSc CSIT entrance · CMAT · 2012), CMAT 2011 Past Question (BSc CSIT entrance · CMAT · 2011) and CMAT 2010 Past Question (BSc CSIT entrance · CMAT · 2010). Answers are written for this site; check them against your class notes.