 # Special Quant Quiz (Arithmetic) SSC CGL Tier-II 2016

Q1. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:
(a) 15
(b) 16
(c)  18
(d) 25
1. Ans.(a)
Sol.
Selling price of x article = cost price of 20 article
Selling price of x article – cost price of x article
= cost price of (20 – x) article
Profit = cost price of (20 – x) article
% profit =(Cost price of (20-x)  article )/(Cost of 20 article)
25%=(20 – x)/20
=1/4=(20 – x)/20
20 = 80 – 4x
4x = 60

x = 15

Q2. A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?
(a) Rs. 18.20
(b) Rs. 70
(c) Rs. 72
(d) Rs. 88.25
2. Ans.(c)
Sol. Let the cost price = 100x
Sell price = 122.5x
According to question,
122.5x = 392
x=392/122.5=3.2
100x = 320

Profit = 392 – 320 = 72

Q3. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
(a) Rs. 45
(b) Rs. 50
(c)  Rs. 55
(d)  Rs. 60
3. Ans.(d)
Sol. Selling of 17 ball = cost price of 17 – cost price of 5 ball
Selling of 17 ball = cost price of 12 ball
Cost price of 12 ball = 720

Cost price of ball =720/12=60

Q4. A series of discounts of 30%, 20% and 10% is equivalent to a single discount of ….
(a) 70%
(b) 49.6%
(c) 48.4%
(d) 49.4%
4. Ans.(b)
Sol. x is a mark price
Selling price =x(1-30/100)(1-20/100)(1-10/100)
= x×.7×.8×.9
= .504x
Single discount % =((x – .504x))/x×100

⇒ 49.6%

Q5. Mohan Sells 25 toys and gains the S.P. of 9 toys. Find his percentage profit.
(a) 56.2%
(b) 60%
(c) 40%
(d) 56.4%
5. Ans.(a)
Sol.  Cost price of 25 toys = sell price of 16 toys
(Sell price)/(Cost price )=25/16
% profit =(SP/CP-1)×100
=9/16×100

= 56.2%

Q6. If two numbers are in the ratio 6 :13 and their least common multiple is 312, then what is the sum of numbers?
(a) 57
(b) 67
(c) 75
(d) 76
6. Ans.(d)
Sol.
Let the first number = 6x
Second number = 13x
LCM of first and second number = 6 × 13 × x
6 × 13x = 312
x=312/(6×13)
x = 4

Sum of number = 6 × 4 + 13 × 4 = 76

Q7. When n is divided by 4, the remainder is 3. What is the remainder when 2n is divided by 4?
(a) 1
(b) 3
(c) 2
(d) None of these
7. Ans.(c)
Sol.
n = 4x + 3
2n = 2(4x + 3)
2n = 8x + 6
According to question,
Remainder =(8x + 6)/4

remainder = 2

Q8. Find the smallest number required to be added to 2000 to get a number exactly divisible by 17.
(a) 6
(b) 4
(c) 2
(d) 3
8. Ans.(a)
Sol.
When 2000 is divided by 17 remainder is equals to 11

Smallest number required to be added = 17 – 11 = 6

Q9. Two numbers are in the ratio 3: 4, and their LCM is 108. Find their HCF.
(a) 9
(b) 8
(c) 6
(d) 5
9. Ans.(a)
Sol.
Let the first number = 3x
Second number = 4x
LCM of these two number = 3 × 4x = 12x
12x = 108
x = 9
HCF of these two number = x

= 9

Q10. A number P4571203K is divisible by 18. Which of the following values can P and K take?
(a) 1, 2
(b) 2, 3
(c) 3, 3
(d) 6, 8
10. Ans.(d)
Sol.
Number = P4571203K
If the number is divisible by 18, then it should be divisible by 2 and 9
So, P + 4 + 5 + 7 + 1 + 2 + 0 + 3 + k = 22 + P + K
⇒ P + K = 5 or P + K = 14
And K must be even number or zero

So, from the following option (d) option fulfilled the condition.

Q11. Six-elevenths of a number is equal to 22% of second number. Second number is equal to the one-fourth of third number. The value of the third number is 2400. What is the 45% of the first number?
(a) 109.8
(b) 111.7
(c) 117.6.0
(d) None of these
11. Ans.(d)
Sol.
Third number = 2400
Second number =1/4×2400=600
First number ×6/11=22% of 600
First number =11/6×600×22/100=242

45% of the first number =(45 × 242)/100=108.90

Q12. In an entrance examination Ritu scored 56% marks, Smita scored 92% marks and Rina scored 634 marks. The maximum marks of the examination are 875, what are the average mark scored by all the three girls together:
(a) 929
(b) 815
(c) 690
(d) 643
12. Ans.(d)
Sol.
Maximum marks in examination = 875
Ritu’s marks =875×56/100=490
And Smita’s marks =875×92/100=805
Hence, required average marks =(490 +805 + 634)/3

= 643

Q13. If the denominator of a fraction is increased by 350%, and numerator is increased by 150%, then the resultant fraction is 25/51, what is the original fraction:

13. Ans.(c)
Sol.
The original fraction =25/51 ((100 + 350)/(100 + 150))=25/51×45/25=15/17

Q14.In An election there were only two candidates, one of the candidates secured 40% of votes and is defeated by the other candidate by 298 votes. The total number of votes polled is:
(a) 745
(b) 1460
(c) 1490
(d) 1500
14. Ans.(c)
Sol.
Let the total number of votes polled be x%
Now, according to the question
x×((60-40)/100)=298

x=298×5=1490

Q15. In a test consisting of 80 questions carrying one mark each, Arpita answer 65% of the first 40 question correctly, what percent of the other 40 question does she need to answer correctly to score 75% on the entire test:
(a) 60
(b) 85
(c) 75
(d) 40
15. Ans.(b)
Sol.

⇒(60-40 × 65/100  )/40× 100=34/40×100=85%

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