Quant Quiz (Arithmetic) for SSC CGL Tier-II

Q1. A number is doubled and 6 is added to it. The resultant so obtained is multiplied by four, it becomes 96. What is the number?  
(a) 8
(b) 9
(c) 12
(d) 13
S1. Ans.(b)
Sol. Let the number = x
According to question,
4(2x + 6) = 96
2x + 6 = 24
⇒ x = 9

Q2. When 1 is added to both the numerator and denominator of certain fraction it becomes half and when 1 is subtracted from both the numerator and denominator it becomes 1/3. Find the original fraction.  
(a) 5/9
(b) 4/9
(c) 4/7
(d) 3/7

Q3. There is a vessel holding 40 litres of milk. Initially, 4 litres of milk is taken out form the vessel and 4 litres of water is poured in. After this, 5 litres of mixture is replaced with 5 litres of water. And finally, 6 litres of mixture is replaced with 6 litres of water. How much of the milk (in litres) is there in the vessel now?  
(a) 26.775
(b) 29.16
(c) 24.72
(d) 27.42
S3. Ans.(a)
Sol. Milk in a vessel = 40 litres
After first process
Milk =40×36/40 = 36 litres
Milk after the five litre of mixture replaced and water added =36×(1-5/40) =(36 × 35)/40
Milk after 6 litre of mixture replaced and water added =36×35/40×34/40
Now,
Quantity of Milk in vessel = 26.775

Q4. Having scored 98 runs in the 19th inning, a cricketer increases his average score by 4. What will be his average score after 19 innings?  
(a) 28
(b) 26
(c) 24
(d) 22
S4. Ans.(b)
Sol. Let the average score of the 1st 18 innings be n
18 n + 98 = 19(n + 4) ⇒ n = 22
So, average score after 19th innings = x + 14 = 26.

Q15. Left pan of a faulty balance weighs 100 grams more than its right pan. A shopkeeper keeps the weight measure in the left pan while buying goods but keeps it in the right pan while selling his goods. He uses only 1 kg weight measure. If he sells his goods at cost price, what is his gain percentage?  
(a) 200/11%
(b) 100/11%
(c) 1000/9%
(d) 200/9%
S15. Ans.(d)
Sol. Assume cp/1000 gm = Rs. 1000
He gets 1100 g for Rs. 1000 and
Sells 900 g for Rs. 1000.
So, cp/gram 1000/1100= Rs. 0.90/g
Sp/gram =1000/900 = Rs. 1.11/g
So, profit% =(1.11 – 0.90)/0.9×100
Hence, option (d) is the answer.

Q6. Mohan bought 3 cows for Rs. 21,200. He sells all the three cows for Rs. 22,360. On the first cow, he makes a profit of Rs. 560 at 10% on cost price. On the second, he makes a profit of Rs. 1440 at 20% on cost price. Find the % profit of % loss made on the third cow.  
(a) 20% profit
(b) 20% loss
(c) 10% profit
(d) 10% loss
S6. Ans.(d)
Sol. CP of first cow = 560(100/10) = Rs. 5600
CP of second cow =1440(100/20)= Rs. 7200
⇒ CP of third cow = 21,200 – (5600 + 7200) = Rs. 8400
Now, total profit = Rs. (22,360 – 21,200) = Rs. 1160
⇒ Profit made on 3rd cow = 1160 – (560 + 1440) = -840
= Rs. 840 loss.
So, % loss =840/8400×100=10%

Q7. A seller gives a discount of 20% on a pen and sells it at Rs. 960. Had he not given the discount, profit would have been 40%. If he wants to gains a profit of 54% on the cost price of watch, what should be selling price?  
(a) Rs. 1320
(b) Rs. 1400
(c) Rs. 1200
(d) Rs. 1280
S7. Ans.(a) 
Sol. Mark price of pen =960/80=1200
C.P. of pen =1200/1.4
To gain a profit of 54% selling price =1200/1.4×1.54=1320

Q8. A dishonest dealer professes to sell his goods at a profit of 20% and also uses weight 800 gm at the place of 1 kg. Find actual gain percentage.  
(a) 40%
(b) 45%
(c) 50%
(d) 60%
S8. Ans.(c)
Sol. Let the price of each gram goods is 1 Rs
Selling price of 1000 gm good = 1000 × 1.2 = 1200 Rs.
Cost price = 800 Rs.
Profit =((S.P.)/(C.P.)-1)×100 = 50%

Q9. The ratio of the age of a man and his wife is 4 : 3. After 4 years, this ratio will be 9:7. If at the time of the marriage, the ratio was 5:3, then how many years ago they were married?   
(a) 12 years
(b) 8 years
(c) 10 years
(d) 15 years

Q10. What least number must be subtracted from each of the numbers 14, 17, 34 and 42 so that the numbers left are proportional?  
(a) 0
(b) 1
(c) 2
(d) 7
S10. Ans.(c)
Sol. From the given option,
If we subtract 2 from each number, then number will become 12, 15, 32 and 40
⇒ 12/15=32/40
Hence, option (c) is a right option.

Q11. A’s efficiency is 50% of efficiency of B and C together and efficiency of A and C is 19/8 times the efficiency of B. Ratio of efficiency of B and C is 4:5. If all of them working together complete the work in 13 1/3 days. In how many days B can do the work, working alone?  
(a) 45 days
(b) 50 days
(c) 30 days
(d) 50 days

Q12. Two workers earned Rs. 225 first worked for 10 days and the second for 9 days. How much did each of them get daily if the first worker got Rs. 15 more for working 5 days than the second worker got for working 3 days?  
(a) Rs. 11.70, Rs. 12.00
(b) Rs. 10.80, Rs. 13.00
(c) Rs. 11.25, Rs. 12.50
(d) Rs. 12.60, Rs. 11.00
S12. Ans.(b)
Sol. Let A got Rs x per day and B got Rs y per day.
So, 10x + 9y = 225 and 5x = 3y + 15
⇒ x = 10.80, y = 13

Q13. A boat travels downstream to cover a distance of 80 km in 5 hrs and upstream to cover a distance of 48 km in 8 h. Find the speed of boat in still water.  
(a) 11 km/h
(b) 12 km/h
(c) 14 km/h
(d) 16 km/h
S13. Ans.(a)
Sol. Downstream speed = 80/5 = 16 kmph = Speed of boat in still water + Speed of current
Upstream speed = 48/8 = 6 kmph = Speed of boat in still water – Speed of current
So, speed of boat in still water = (Downstream speed + Upstream speed)/2 = 11 kmph

Q14. Forty per cent of employees of a certain company are man, and 75% of the men earn more than Rs. 25,000 per year. If 45% of the company’s employees earn more than Rs. 25,000 per year, what fraction of the women employed by the company earn less than or equal to Rs. 25,000 per year?  
(a) 2/11
(b) 1/4
(c) 1/3
(d) 3/4
S14. Ans.(d)
Sol. Men women
40% 60%
Out of 40% men, 75% earn more than Rs. 25,000. Hence, 30% of the company (men) earn more than Rs. 25,000.
But in all 45% of the employees earn more than Rs. 25,000.
Hence among women 15% earn more than Rs. 25,000 and the remaining (60 – 15) % earn less than or equal to Rs. 25,000.
Therefore, the fraction of women =45/60=3/4

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