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Quant Quiz “Arithmetic ” for SSC CGL 2016

Quant Quiz "Arithmetic " for SSC CGL 2016_40.1

Q1. A path of uniform width surrounds a
circular park. The difference of internal and external circumferences of this
circular path is 132 metres. Its width is: (Take π=22/7) 
(a) 22m
(b) 20 m
(c) 21m
(d) 24m

Sol. Let the internal radius of the park be r and the external
radius (with the path) be R. 
The difference between the internal and external circumferences is
132m. 
so 2πR-2πr = 132 
 =>2π(R-r) = 132
=R – r = 132/2π 
= (132 * 7)/ (2 * 22) = 21 
Hence, the width of path 
= 21 meters 
Q2. A is thrice as good as workman as B
and therefore, able to finish a job in 60 days less than B. Working together
they will do it in: 
(a) 120/7 days 
(b) 90/4days 
(c) 25 days 
(d) 30 days

Sol.  Let A can finish the whole work in x days
so B can finish the same work in 3x days.
According to question.
3x – x = 60
so x = 30 days
so A can finish the work in 30 days and B can finish the same work in 90 days.
so (A + B)’s one day’s work = 1/30 + 1/90 = 4/90
so (A + B)’s can finish the complete work in = 90/4 days.
Q3. A man, a woman and a boy can
complete a work in 20 days. 30 days and 60 days respectively. How many boys
must assist 2 men and 8 women so as to complete the work in 2 days? 
(a) 8 
(b) 12 
(c) 4 
(d) 6 

Sol. Part of work done by 2 men and 8 women in 2 days.
= 2(2/20 + 8/30)
= 2(1/10 + 8/20) = 2((3+8)/30)
= 22/30 = 11/15
Remaining work = 1 – 11/15 = 4/15
work done by 1 boy in 2 days
= 2/60 = 1/30
Number of boys done 4/15 work = 4*30/15
                     = 8 boys

Q4. 8% of the voters in an election did
not cast their votes. In this election, there were only two candidates. The
winner by obtaining 48% of the total votes defeated his contestant by 1100
votes. The total number of voters in the election was:
(a) 21000
(b) 23500
(c) 22000
(d) 27500

Sol. Let the total number of votes be 100. 
Number of cast votes = 100-8 = 92 
so Number of votes obtained by the winner = 48 
Number of votes obtained by the loser = 92 – 48 = 44 
If the difference of win be 4 votes, total voters = 100 
so When the difference be 1100 votes, total voters 
= 100/4 * 1100 = 27500
Q5. A, B and C start at the same time in
the same direction to run around a circular stadium. A completes a round in 252
seconds, B in 308 seconds and C in 198 seconds, all starting at the same point.
After what time will they next meet at the starting point again?
(a) 46 minutes 12 seconds
(b) 45 minutes 12 seconds
(c) 42 minutes 36 seconds
(d) 26 minutes 18 seconds

Sol.  Required time = LCM of 252. 
308 and 198 seconds. 
Now, 252 = 2 * 2* 3 * 3 * 7 
308 = 2 * 2 * 7 * 11 
198 = 2 * 3 * 3 * 11 
so LCM = 2 * 2* 3 * 3 * 7 * 11 
= 36 * 77 seconds 
= (36 * 77)/60 minutes 
= 231/5 = 46 minutes 12 seconds

Q6. A
merchant marks his goods up by 60% and then offers a discount on the marked
price. If the final selling price after the discount results in the merchant
making no profit or loss, what was the percentage discount offered by the
merchant?
(a) 60
(b) 40
(c) 37.5
(d) Depends on the cost price 


Sol. Let C.P = 100
Markup price = 100 ×1.6
                      = 160
S.P. = C.P.
Discount % = (60/160) × 100
                  = 37.5%
Q7. A trader
makes a profit equal to the selling price of 75 articles when he sold 100 of
the articles. What percentage of profit did he make in the transaction?
(a) 33.33%
(b) 75%
(c) 300%
(d) 150%


Sol. Let S be the Selling price of 1 article,
Therefore, the selling price of 100 articles=100 S
Profit earned by selling these 100 articles=selling price of 75 articles = 75 S
We know that, Selling Price (SP)=Cost Price (CP)+Profit
So, we get
100S = CP+75S. Hence, CP
=100S-75S=25S
Profit percentage = Profit/(Cost Price)×100%

=75S/25S×100%=300% 

Q8. A sum amounts to Rs. 98010 in 2 years and to
Rs. 107811 in 3 years compounded annually. What is the sum?

(a) Rs. 81000
(b) Rs. 84000
(c) Rs. 87000
(d) Rs. 90000

Sol. P
(1 + R/100) ^3 = 107811

P (1 + r/100) ^2 = 98010

1 + r/100 = 107811/98010
r/100 = 9801/98010
r = 10%

P × (1 + 10/100) ^2 = 98010
P = 98010 * 100/121 = 81000


Q9. The simple interest on a certain sum
for 8 months at 4% per annum is Rs. 129 less than the simple interest on the
same sum for 15 months at 5% per annum. The sum is:
(a) Rs. 2,580
(b) Rs. 2400
(c) Rs. 2,529
(d) Rs. 3600

Sol. Let the sum be Rs. x 
(x * 5 * 15)/(100 * 12) – (x * 4 * 8)/(100 * 12) = 129 
= x/(100 * 12) (75 – 32) = 129 
X = (129 * 1200)/ 43 
Rs.   3600 
Q10.Compound interest (compounded annually) on a certain sum of money for
2 years at 4% per annum is Rs. 102. The simple interest on the same sum for the
same rate and for the same period will be: 
(a) Rs. 99   
(b) Rs. 101  
(c) Rs. 100  

(d) Rs. 98


Sol.
Let principal be Rs. P

According to question.
102 = p {(1+r/100) ^t- 1}
Where r = 4% and t
= 2 years
P = 102/ ((26/25) ^2- 1) =Rs.1250 
so Simple Interest at rate of 4% for 2 years
= (1250 ×4 ×2)/100=Rs.100 
Required answer is Rs. 100


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