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# Quant Quiz “Misc.” for SSC CGL 2016 Q1. If a : b = c : d = e : f = 1 : 2, then (3a + 5c + 7e) : (3b + 5d + 7f) = ?
(a) 1 : 2
(b) 1 : 4
(c) 2 : 1
(d) 8 : 7

Q2. In a class, the number of girls is 20% more than that of the boys. The strength of the class is 66. If 4 more girls are admitted to the class, what will be the ratio of the number of boys to that of the girls ?
(a) 1 : 2
(b) 1 : 4
(c) 3 : 4
(d) 3 : 5

Q3. Three numbers A, B and C are in the ratio 1 : 2 : 3. Their average is 600. If A is increased by 10% and B is decreased by 20%, then to get the average increased by 5%, C will be increased by
(a) 90
(b) 100
(c) 150
(d) 180

Q5. If a/3=b/5=c/7, then find the value of (a+b+c)/b
(a) 1/5
(b) 1/3
(c) 3
(d) 5
S5. Ans.(c)
Sol. Given that a/3=b/5=c/7
⇒ Let a/3=b/5=c/7=K
⇒ a = 3k, b = 5k, c = 7k

=(a + b + c)/b=(3 + 5 + 7)K/5K=15/5=3

Q6. If the average of m numbers is n^2 and that of n number is m^2, then the average of (m + n) number is :
(a) m + n
(b) mn
(c) m/n
(d) m – n
S6. Ans.(b)
Sol. Total of (m + n) numbers =(mn^2+nm^2 )
= mn(m + n )
Average of (m + n) numbers

=(mn (m+n))/((m+n) )=mn

Q7. The average of four numbers A, B, C, D is 40. The average of four numbers A, B, E, F is also 40. Which of the following must be true ?
(a) A + B = C + D
(b) A – B = C – D
(c) C + D = E + F
(d) C = E and D = F
S7. Ans.(c)
Sol. (A + B + C + D) = (40 × 4) = 160,
(A + B + E + F) = (40 × 4) = 160
∴ A + B + C + D = A + B + E + F

⇒ C + D = E + F

Q8. The mean of 100 items was found to be 30. If at the time of calculation, Two  items were wrongly taken as 32 and 12 instead of 23 and 11, the correct mean is :
(a) 29.4
(b) 29.5
(c) 29.8
(d) 29.9
S8. Ans.(d)
Sol. Sum of all the items = (100 × 30) = 3000
Correct sum = (3000 + 23 + 11 – 32 – 12) = 2990

Correct mean =2990/100=29.9

Q9. The average age of 24 boys and their teacher is 15 years. When the teacher’s age is excluded, the average age decreases by 1 year. The age of the teacher is
(a) 38 years
(b) 39 years
(c) 40 years
(d) 41 years
S9. Ans.(b)
Sol. Sum of the ages of 24 boys and 1 teacher = (15 × 25) = 375 years
Sum of ages of 24 boys = (24 × 14) = 336 years

Age of the teacher = (375 – 336) = 39 years

Q10. The mean temperature of Monday to Wednesday was 37° C and that of Tuesday to Thursday was 34° C. If the temperature on Thursday was 34° C. If the temperature on Thursday was 4/5 of that of Monday, then what was the temperature on Thursday ?
(a) 36.5°
(b) 36°
(c) 35.5°
(d) 34°
S10. Ans.(b)
Sol. M + T + W = (37 × 3) = 111 …(i)
T + W + Th = (34 × 3) = 102 …(ii)
Th =4/5  M
⇒ M=4/5  Th  …(iii)
(ii) – (i) gives, M – TH = 9
⇒ 5/4 Th – Th = 9
Let the temperature on Thursday be x°
Then,
5/4 x-x=9
⇒ 5x – 4x = 36
⇒ x = 36°

Hence, the temperature on Thursday was 36°C

Q11. The average age of 8 person in a committee is increased by 2 years when two men aged 35 years and 45 years are substituted by two women. The average age of these two women is
(a) 28 years
(b) 30 years
(c) 36 years
(d) 48 years
S11. Ans.(d)
Sol. Increased age = (2 × 8) years = 16 years
Total ages of two men = (35 + 45) = 80 years
Total ages of new women = (80 + 16) = 96 years

Average age of these two women =96/2 = 48 years

Q13. The average of the largest and smallest 3 digit numbers formed by 0, 2 and 4 would be
(a) 213
(b) 303
(c) 312
(d) 222
S13. Ans.(c)
Sol.

(204 + 420)/2=624/2=312

Q14. Out of nine persons, 8 persons spent Rs. 30 each for their meals. The ninth one spent Rs. 20 more than the average expenditure of all the nine. The total money spent by all of them was
(a) Rs. 260
(b) Rs. 290
(c) Rs. 292.50
(d) Rs. 400.50
S14. Ans.(c)
Sol. Expenditure of 9th person = Rs. x
∴x-(x + 8 × 30)/9=20
∴(9x – x – 240)/9=20
⇒ 8x – 240 = 180
⇒ 8x = 240 + 180 = 420
⇒ x=420/8=52.5
Total expenditure = 52.5 + 240 = Rs. 292.5

Q15. Members of a team are weighed consecutively and their average weight calculated after eafh  member is weighed. If the average weight increases by one kg each time, how much heavier is the last player than the first one ?
(a) 4 kg
(b) 20 kg
(c) 8 kg
(d) 5 kg

S15. Ans.(c)
Sol. Weight of first member = x kg
Wight of second number = (x + 2) kg
Weight of fifth member = (x + 8) kg
∴ Difference = x + 8 – x = 8 kg