Quantitative Ability

(d) Rs. 40,000

Let the original salary be Rs. $x$.

After 1 year: $1.1x$
After 2 years: $1.1 \times 1.1x = 1.21x$

This becomes Rs. 48,400 in the third year, so:

$1.21x = 48400$

$x = \frac{48400}{1.21} = 40000$

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

(c) Rs. 12,484

Using (A=P(1+r)^n): [ 15660=P(1+0.12)^2=P(1.12)^2 ] [ P=\frac{15660}{1.2544}=12484 \text{ (approximately)} ] So the sum lent is Rs. 12,484.

(c) 10%

For simple interest, doubling means interest = principal: $P\times \frac{R\times 10}{100}=P \Rightarrow R=10\%$

(a) 2

$1056 \div 23$ leaves remainder $21$ since $23 \times 45 = 1035$. To reach the next multiple of 23, add $23 - 21 = 2$.

(b) Rs. 4000

Extra interest over 3 years at 4% higher rate: $P \times \frac{4}{100} \times 3 = 480$ $0.12P = 480 \Rightarrow P = 4000$

(c) 40

Let current ages be Usha $=U$, Nisha $=N$, Asha $=A$.

When Usha was thrice Nisha, Asha was 25: $U-t_1=3(N-t_1),\quad A-t_1=25$

When Nisha was half Asha, Usha was 34: $N-t_2=\frac12(A-t_2),\quad U-t_2=34$

Also, $U+N+A=100$

Solving gives: $N=22,\quad A=38,\quad U=40$

So Usha is 40 years old.

(d) 20

Total present age of 5 members = (5 \times 24 = 120) years.
The youngest is 8 years old, so 8 years ago the other 4 members’ total age was:

[ 120 - 8 - (4 \times 8) = 120 - 8 - 32 = 80 ]

Average of those 4 members at that time:

[ \frac{80}{4} = 20 ]

So the average age was 20 years.

(a) 1951609

$1397^2 = (1400-3)^2 = 1400^2 - 2\cdot1400\cdot3 + 3^2 = 1960000 - 8400 + 9 = 1951609$.

(b) 22

Old average = 14. After taking 5 wickets for 26 runs, average improves by 2, so new average = 12.

Let old wickets = $W$ and old runs = $R$.

$\frac{R}{W}=14 \Rightarrow R=14W$

After the final match:

$\frac{R+26}{W+5}=12$

Substitute $R=14W$:

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

$14W+26=12W+60$

$2W=34 \Rightarrow W=17$

Total wickets after final match:

$W+5=17+5=22$

So the answer is (b) 22.

(c) 27

Correct total marks = $75n$.
Wrong total marks = $75n - (96-69)=75n-27$.
Decrease in average = $\frac{27}{n}=1$, so $n=27$.

(d) 75

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

(c) 11

Let the number of subjects be $n$.
Total marks initially $=63n$.
With 20 more in Geography and 2 more in History, total increases by $22$.

$65=\frac{63n+22}{n}$ $65n=63n+22$ $2n=22 \Rightarrow n=11$

So, there were 11 subjects.

(a) $2x=y+z$

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

(b) 2700

Valid votes = $80\%$ of 7500 = 6000.
The other candidate got (100% - 55% = 45%) of valid votes = $45\%$ of 6000 = 2700.

(c) 5%

The final population printed, 2652500, appears to be a typo for 265250. Only that value gives one of the listed options, so it is assumed here.

Increase over the decade $= 265250 - 175000 = 90250$.

$\text{Percent increase} = \frac{90250}{175000}\times 100 \approx 51.57\%$

$\text{Average per year} = \frac{51.57}{10} \approx 5.16\% \approx 5\%$

(Taken literally, 2652500 would give about 141.6% per year, which matches no option.)

(d) 7.2

Speed $= \frac{600}{5} = 120$ m/min $= 120 \times 60 = 7200$ m/h $= 7.2$ km/h.

(c) Rs. 300

Let the number of children be $n$. Then the sum of money is $60n$.

When one more child joins, there are $n+1$ children and each receives Rs. 50, so the same sum is $50(n+1)$.

$60n = 50(n+1)$ $60n = 50n + 50$ $10n = 50 \Rightarrow n = 5$

Sum of money $= 60 \times 5 =$ Rs. 300

(c) 5

Cost price per banana = $\frac{1}{6}$ rupee.
To gain 20%, selling price per banana $= \frac{1}{6} \times 1.20 = \frac{1}{5}$ rupee.
So, bananas sold for 1 rupee $= 5$.

(a) 9944

The largest 4-digit number is 9999. Dividing by 88 gives remainder 55, so subtract 55:
$9999 - 55 = 9944$, which is exactly divisible by 88.

(b) 5184

Worked literally, $(12)^3 \times 6 + 432 = 1728 \times 6 + 432 = 10368 + 432 = 10800$, which matches none of the options, so the expression is misprinted in the source. The only option that fits the numbers is $5184 = 12^3 \times 3 = 1728 \times 3$, so (b) is the intended key.

(b) 4991

Required total for 6 months $= 6500 \times 6 = 39000$.
Sum of first 5 months $= 6435+6927+6855+7230+6562=34009$.
Sixth month sale $=39000-34009=4991$.

(c) 285

A 30-day month starting on Sunday has 5 Sundays and 25 other days.
Average = $\dfrac{5 \times 510 + 25 \times 240}{30} = \dfrac{2550 + 6000}{30} = 285$ visitors per day.

(d) Three

The digits in ascending order are 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.

(b) 10

Let the number be $x$.
$24\%$ of $x = 6$

$0.24x = 6 \Rightarrow x = \frac{6}{0.24} = 25$

Now, $40\%$ of $25$:

$0.40 \times 25 = 10$

(c) 75

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

(b) 10010

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

(c) 361

Since $361^2 = 130321$, the square root of 130321 is 361.

(c) Rs. 15

Dividend per share = 9% of Rs 20 = Rs 1.80.
For a 12% return, market value = $\dfrac{1.80}{0.12} = Rs\ 15$.

(d) 64

The pattern is $n^2+1$:
$1^2+1=2,\ 2^2+1=5,\ 3^2+1=10,\ 4^2+1=17,\ 5^2+1=26,\ 6^2+1=37,\ 7^2+1=50$.
The next term should be $8^2+1=65$, not 64. So 64 is the odd one out.

(c) 54

All the given numbers except 54 are divisible by 5. Hence, 54 is the odd one out.

(c) 14

All the given numbers are odd except 14, which is even. So, 14 is the odd one out.

(b) 23
All the other numbers are perfect squares: $1=1^2$, $4=2^2$, $9=3^2$, $16=4^2$, $25=5^2$, $36=6^2$, while 23 is not a perfect square.

(c) 29

29 is the only prime number among the given options; 26, 28 and 30 are even composite numbers.

(a) 108

Total runs in 10 innings = $108 \times 10 = 1080$
After next innings = $1080 + 110 = 1190$
New average = $\frac{1190}{11} = 108.18$ runs

The exact new average is 108.18, so among the given options the closest is (a) 108.
(If the question expects an exact value, none of the options is correct.)

(a) 6.25

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

(c) 40

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

(c) Rs. 698

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

(a) 625

For 2 years, the difference between CI and SI is
$P\left(\frac{r}{100}\right)^2 = 1$ $P\left(\frac{4}{100}\right)^2 = 1 \Rightarrow P \times 0.0016 = 1 \Rightarrow P = \text{Rs. }625$

None of the given options.
$\frac{2}{6}\div\frac{5}{7}=\frac{1}{3}\times\frac{7}{5}=\frac{7}{15}$, which is not listed among (a)–(d).

(d) 1/125

$5^{-3} = \dfrac{1}{5^{3}} = \dfrac{1}{125}$

(Note: the question contains a typo — $5^{5} = 3125$, while $5^{3} = 125$. Either way, $5^{-3} = \dfrac{1}{125}$.)

(b) 135

Room area $=9 \times 10 = 90$ sq ft.
Tile width $=8$ inches $=\frac{2}{3}$ ft, length $=1$ ft.
Tile area $=\frac{2}{3} \times 1=\frac{2}{3}$ sq ft.
Number of tiles $=90 \div \frac{2}{3}=90 \times \frac{3}{2}=135$.

(a) 2 : 3 : 4

Increased seats:
Mathematics $=5\times1.40=7$
Computer $=7\times1.50=10.5$
Accounting $=8\times1.75=14$

Ratio $=7:10.5:14=14:21:28=2:3:4$.

(b) 10

The pattern alternates: add 3, then subtract 2. So after 12, subtract 2 to get 10.

None of the options is correct. Solving gives

$6x + 5 = x + 40 \Rightarrow 5x = 35 \Rightarrow x = 7$

Since $x = 7$ does not appear among (a)-(d), the question has no correct option.

(a) -3

Rewrite in slope-intercept form: $y = -3x + 4$, so the coefficient of $x$ gives slope $m = -3$.

(d) 0, 2 and 5

Since (f(x)=0) has solutions (x=-2,0,3), for (f(x-2)=0) we set (x-2=-2,0,3), giving (x=0,2,5).

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

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

(d) $\ln t$

The points are ((e^t,,t)). Let (x=e^t); then (t=\ln x). So the graph is (y=\ln x), i.e. (y=\ln t) when the independent variable is written as (t).

(b) $z^{-1}$

Using the law of indices $a^m \cdot a^n = a^{m+n}$: $z^2 \cdot z^{-3} = z^{2+(-3)} = z^{-1}$

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

Sum $=5\times10=50$, and the given option sums to $50$. Its sorted form is $6,6,9,14,15$, so the median is $9$ and the mode is $6$.

(b) 1.2 cm

Water displaced: $\pi(2)^2(0.6)=2.4\pi$.
So, $\frac{4}{3}\pi r^3=2.4\pi \Rightarrow r^3=1.8 \Rightarrow r\approx1.216$ cm, nearest tenth = 1.2 cm.

(a) 0.057

Probability a device functions properly $=1-0.1=0.9$.
For exactly 7 of 10 devices functioning properly:

$P(X=7)=\binom{10}{7}(0.9)^7(0.1)^3$

$=120 \times 0.4782969 \times 0.001 \approx 0.0574 \approx 0.057$

(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 = $\frac{7}{21} = \frac{1}{3}$.

(c) 20

The sequence should be perfect squares:
$1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2 = 1, 4, 9, 16, 25, 36, 49$.

The term $20$ is not a perfect square; it should be $25$. So the odd man out is $20$, which is option (c).

(d) 2100

Number of ways = $\binom{5}{2} \times \binom{10}{4} = 10 \times 210 = 2100$.