Quant (Arithmetic Misc) Quiz for SSC CGL Tier-II 2016



Directions (1-5): The bar chart given below shows the percentage distribution of the production of various models of a mobile manufacturing company in 2007 and 2008. The total production in 2007 was 35 Lakh mobile phones and in 2008 the production was 44 Lakhs. Study the chart and answer the following questions. 
Percentage of six different types of mobiles manufactured by company over two years

Q1. Total number of mobiles of models A, B and E manufactured in 2007 was    
(a) 24,50,000
(b) 22,75,000
(c) 21,00,000
(d) 19,25,000
S1. Ans.(c)
Sol.  Required answer =(35×30)/100+(35×15)/100+(35×15)/100
=35/100 (30+15+15)=(35×60)/100 = 21 Lakhs

Q2. For which models was the percentage variation in production from 2007 to 2008 the maximum?
(a) B and C
(b) C and D
(c) D and E
(d) A and B
S2. Ans.(d)
Sol. Percentage variation
Model A =(40-30)/30×100=33 1/3
Model B =(20-15)/15×100=33 1/3
Model C =(15-20)/20×100=-25

Q3. What was the difference in the number of B type mobiles produced in 2007 and 2008? 
(a) 3,55,000
(b) 2,70,000
(c) 2,25,000
(d) 1,75,000
S3. Ans.(a)
Sol. Required difference
=(44×20)/100-(35×15)/100
=(880-525)/100=355/100 lakhs = 355000

Q4. If the percentage production of A type mobiles in 2008 was same as that in 2007, then the number of A type mobiles produced in 2008 would have been  
(a) 14,00,000
(b) 13,20,000
(c) 11,70,000
(d) 10,50,000
S4. Ans.(b)
Sol. Required production
=(44×30)/100 Lakhs
= 1320000

Q5. If 85% of the D type mobiles produced in each year were sold by the company, how many D type mobiles remained unsold? 
(a) 76,500
(b) 93,500
(c) 1,18,500
(d) 1,22,500
S5. Ans.(c)
Sol. Required answer
=(35×10/100×15/100+44×10/100×15/100) Lakhs
=(150/10000×79) = 1.1850 Lakhs
= 118500

Q6. The ratio of the adjacent angles of a parallelogram is 7 : 8. Also, the ratio of the angles of quadrilateral is 5:6:7:12. What is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral?  
(a) 168°
(b) 228°
(c) 156°
(d) 224°
S6. Ans.(a)
Sol. Sum of the adjacent angles of a parallelogram is 180°
Smaller angle of the parallelogram =7/15×180=84°
Second largest angle of the quadrilateral =7/30×360=84°
∴ Required sum = 84 + 84 = 168°  

Q7. The difference between the simple and compound interest on a certain sum of money at 5% rate of interest per annum for 2 years is Rs. 15. The sum is:  
(a) Rs. 6500
(b) Rs. 5500
(c) Rs. 6000
(d) Rs. 7000

Q8. Two types of steel are available. Type A contains 5% of nickel and type B contain 40% of nickel. How much of each type be mixed to obtain 140 tonnes of steel containing 30% of nickel?  (a) 40, 100 tonnes
(b) 50, 125 tonnes
(c) 30, 75 tonnes
(d) 60, 180 tonnes

Q9. There are two containers of equal capacity. The ratio of milk to water in the first container is 3:1, in the second container 5:2. If they are mixed up, the ratio of milk to water in the mixture will be:   
(a) 28:41

(b) 41:28
(c) 15:41
(d) 41:15
 S9. Ans.(d)

Sol. Let the capacity of each container be x litres. 
 In first container,
Milk =3x/4 litres
Water =x/4 litres
 In second container
Milk =5x/7 litres
Water =2x/7 litres 
On mixing both, we have,
Quantity of milk =3x/4+5x/7
 =(21x + 20x)/28
=41x/7 litres 
 Quantity of water =x/4+2x/7=(7x + 8x)/28 litres
 =15x/18 litres
 ∴ Required ratio =41x/28:15x/28=41:15 
Q10. There was one mess for 30 boarders in a certain hostel. On the number of boarders being increased by 10, the expenses of the mess were increased by Rs. 40 per month while the average expenditure per head diminished by Rs. 2. Find out actual monthly expenses.  
(a) Rs. 390
(b) Rs. 410
(c) Rs. 360
(d) Rs. 480
S10. Ans.(c) 
Sol. Let Rs. x be the average expenditure for 30 boarders.
∴ 30x + 40 = (x – 2) × 40
Or, x = 12
Hence, actual expenditure = Rs. 12 × 30 = Rs. 360


(a) 2/3
(b) 1
(c) 4/3
(d) 3/4

Q12. If the square root of 5 is 2.236, then the square root of 80 equals = 2.236 times of:  
(a) 2
(b) 2.5
(c) 4
(d) 5
S12. Ans.(c)
Sol. √80=√(16×5)=4√5=4× 2.236


Q13. A number, when divided successively by 4, 5 and 6, leaves remainders 2, 3 and 4 respectively. The least such number is:  

(a) 50
(b) 53
(c) 58
(d) 214
S13. Ans.(c)
Sol. ∴ 4 – 2 = 5 – 3 = 6 – 4 = 2
Now, L∴ Required number = 60 – difference = 60 – 2 = 58
(L.C.M. of 4, 5, 6 = 60)

Q14. While solving a mathematical problem, Samidha squared a given number and then subtracted 25 from it rather than doing what was required she first subtracting 25 from the number and then squaring it. But she got the right answer. What was the given number:?  
(a) 48
(b) 13
(c) 38
(d) Cannot be determined
S14. Ans.(b)
Sol. Suppose the number is x
Then,
x^2-25=(x-25)^2
Or, x^2-25=x^2-50x+625
Or, 50x = 650
∴ x = 13.

Q15. A 2-digit number is 3 times the sum of its digits. If 45 is added to the number, its digits are interchanged. The sum of the digits of the number is: 
(a) 11
(b) 9
(c) 7
(d) 5
S15. Ans.(b)
Sol. Let the two-digit number = xy
According to question,
10x + y = 3(x + y)
7x = 2y …(i)
Again,
10x + y + 45 = 10y + x
y – x = 5 …(ii)
By solving these two equations
x = 2, y = 7
Sum of digits of number = 2 + 7 = 9

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