Quantitative Ability

(b) 16, 28, 36 Let the ages be $4k, 7k, 9k$. Their sum is $20k$. Eight years ago the sum was $20k - 3\times 8 = 20k - 24 = 56$, so $k = 4$. Ages: $16, 28, 36$.

(c) 64 Total ways: $\binom{9}{3} = 84$. Ways with no black ball: $\binom{6}{3} = 20$. At least one black: $84 - 20 = 64$.

(c) $\sqrt{12}\times\sqrt{3}=\sqrt{12\times 3}$ For (c), $\sqrt{a}\sqrt{b}=\sqrt{ab}$ holds, giving $\sqrt{36}=6$. Statement (a) is false ($4+\sqrt3 \ne \sqrt{19}$) and (b) is false ($\sqrt{15}-\sqrt5 \ne \sqrt{10}$). So (d) and (e) are also wrong.

(d) 8/15 Together, one day of work is $\tfrac{1}{15}+\tfrac{1}{20}=\tfrac{7}{60}$. In 4 days: $\tfrac{28}{60}=\tfrac{7}{15}$. Work left: $1-\tfrac{7}{15}=\tfrac{8}{15}$.

(b) 16 Let the cost price of one article be $C$. Then $20C = x \cdot SP$, where $SP = 1.25C$. So $20C = 1.25\,xC$, giving $x = 16$.

(c) 57% Total votes $=1136+7636+11628=20400$. Winner's share $=\dfrac{11628}{20400}\times 100 = 57\%$.

(b) Rs 1190 Selling price $=1400\times\dfrac{85}{100}=\text{Rs }1190$.

(b) 42 Let the numbers be $n$ and $n+2$. Then $(n+2)^2-n^2=4n+4=84$, so $n=20$. The numbers are 20 and 22, and their sum is $42$.

(b) 42, 33 Let the larger mark be $x$, so the other is $x-9$. Then $x = 0.56(2x-9) = 1.12x - 5.04$, so $0.12x = 5.04$ and $x = 42$. The marks are $42$ and $33$ (check: $42/75 = 56\%$).

(c) $\sqrt{k}<1$ For $0<k<1$, taking square roots preserves the inequality: $0<\sqrt{k}<\sqrt{1}=1$. For example, $k=0.25$ gives $\sqrt{k}=0.5<1$.

(d) A is equal to B

$A=\dfrac{x}{100}\cdot y=\dfrac{xy}{100}$ and $B=\dfrac{y}{100}\cdot x=\dfrac{xy}{100}$, so $A=B$.

(a) 13

Multiply both sides by $(x-2)(x-1)$: $3x+7=A(x-1)-10(x-2)$. Put $x=2$: $13=A(1)$, so $A=13$.

Check at $x=1$: $10=-10(-1)=10$, which holds.

(d) 31 kg

$A+B+C=3\times45=135$, $A+B=80$, $B+C=86$.

$B=(A+B)+(B+C)-(A+B+C)=80+86-135=31$

(b) $10500

Simran invested $50000$ for 36 months and Nanda $80000$ for 30 months, so the profit ratio is $50000\times36:80000\times30=18:24=3:4$.

Simran's share $=\dfrac{3}{7}\times24500=10500$.

(c) 100

Students of 8 years $=48$, so students above 8 years $=\tfrac23\times48=32$. Students below 8 years are $20\%$ of the total $T$, so the other $80\%$ are $48+32=80$.

$0.8T=80\Rightarrow T=100$

(a) 2700

Valid votes $=80\%$ of $7500=6000$. The other candidate got $45\%$ of the valid votes:

$0.45\times6000=2700$

(c) $f(2)$

For $x<2$, $f(1)=1$. For $x\ge2$, $f(2)=2(2)-3=1$. Hence $f(1)=f(2)$.

(c) 18 cm

Let length $l$ and breadth $b$. Then $\dfrac{2(l+b)}{b}=5$, so $2l+2b=5b$ and $l=1.5b$.

Area: $1.5b^2=216\Rightarrow b^2=144\Rightarrow b=12$ cm, so $l=18$ cm.

(b) 720

LEADING has 7 distinct letters, with vowels E, A, I. Treat the three vowels as one block, giving $5$ units (L, D, N, G and the block), which can be arranged in $5!=120$ ways. The vowels can be arranged inside the block in $3!=6$ ways.

$120\times6=720$

(a) 4:5

Let the third number be $x$. The numbers are $1.2x$ and $1.5x$, so the ratio is $1.2:1.5=12:15=4:5$.

(b) 2/7

Total tickets $=10+25=35$.

$P(\text{prize})=\frac{10}{35}=\frac27$

(c) 23

Let the numbers be $a,b$ with $ab=120$ and $a^2+b^2=289$.

$(a+b)^2=a^2+b^2+2ab=289+240=529\Rightarrow a+b=23$

(The numbers are 8 and 15.)

(c) 41

Let $P$ = multiples of 2, $C$ = multiples of 5, $M$ = multiples of 7 among 1 to 120.

$n(P)=60,\ n(C)=24,\ n(M)=17$; $n(P\cap C)=12$ (multiples of 10), $n(P\cap M)=8$ (multiples of 14), $n(C\cap M)=3$ (multiples of 35), $n(P\cap C\cap M)=1$ (70).

$n(P\cup C\cup M)=60+24+17-12-8-3+1=79$

Students opting for none $=120-79=41$.

(c) 30

The meal serves 120 men or 200 children, so 1 child equals $\tfrac{120}{200}=0.6$ man. The 150 children use the meal of $150\times0.6=90$ men. The remaining meal is for $120-90=30$ men.

(c) 0.06

$0.000216=\frac{216}{10^6}\Rightarrow\sqrt[3]{0.000216}=\frac{6}{10^2}=0.06$

(b) 85 kg

Total weight increases by $8\times2.5=20$ kg, so the new person weighs $65+20=85$ kg.

(a) 0

The terms alternate $-1,+1,-1,+1,\dots$. The 10 terms form 5 pairs, each summing to $0$, so the sum is $0$.

(c) 270

Let the smaller number be $s$. Then the larger is $6s+15$ and

$(6s+15)-s=1365\Rightarrow5s=1350\Rightarrow s=270$

(d) $\log x^n=n+\log x$

The correct rule is $\log x^n=n\log x$, so (d) is false. Statements (a), (b) and (c) are true, Option (e) says both (a) and (d) are false, but (a) is true, so (e) is wrong.

(a) 4

For a common remainder, the divisor must divide the differences: $91-43=48$, $183-91=92$, $183-43=140$.

$\gcd(48,92)=4,\quad \gcd(4,140)=4$

So the greatest such number is $4$.

(b) 17.2

$\frac{29.94}{1.45}=\frac{2994/100}{14.5/10}=\frac{2994}{14.5}\times\frac{10}{100}=172\times0.1=17.2$

(d) 8

The positive square root of 16 is $4$, and its cube is $64$. The number whose square is $64$ is $8$.

(c) $x<-\dfrac{23}{5}$

$7x-5>12x+18\Rightarrow-23>5x\Rightarrow x<-\frac{23}{5}$

Dividing by the positive number 5 keeps the inequality direction.

(b) 17

$255=3\times5\times17$, so the prime factors are 3, 5 and 17, and the largest is 17.

(d) 126

Either 5 students go in the larger car and 3 in the smaller, or 4 and 4 (the cars cannot hold more than 8 together, and all 8 must travel).

$\binom85+\binom84=56+70=126$

(e) $x-2z$

Since $x=y+z$, we get $y=x-z$. Then $y-z=(x-z)-z=x-2z$.

(b) 4000

$P+Q=2\times5050=10100$, $Q+R=2\times6250=12500$, $P+R=2\times5200=10400$.

Adding: $2(P+Q+R)=33000$, so $P+Q+R=16500$. Then $P=16500-12500=4000$.

(a) 2

$1056=23\times45+21$, so the remainder is 21. The least number to add is $23-21=2$, giving $1058=23\times46$.

(c) 30 birds

B kills once in 2 shots, so B misses once in every 2 shots. 27 misses means B fired $27\times2=54$ shots. A fires 5 shots for every 3 of B, so A fired $54\times\frac{5}{3}=90$ shots. A kills once in 3 shots, so A killed $90/3=30$ birds.

(a) 11

The relationship $x\le 8$ is false only when $x>8$. Of the options, only 11 exceeds 8; $-5,4,0,6$ all satisfy it.