Get busy working on these arithmetic problems
Hyderabad: This article is in continuation to the last article focusing on the percentage topic. Here are some practice questions along with solutions that will help you in your preparation for the State government recruitment jobs. 1. The TV2022 is a very popular channel. It telecasts programmes from 6 am to 12 noon. It telecasts […]
Hyderabad: This article is in continuation to the last article focusing on the percentage topic. Here are some practice questions along with solutions that will help you in your preparation for the State government recruitment jobs.
1. The TV2022 is a very popular channel. It telecasts programmes from 6 am to 12 noon. It telecasts 40 advertisements each of 10 seconds and 20 advertisements each of 30 seconds. What is the percentage of time devoted in for the advertisements?
a) 1.45% b) 1.54% c) 1.75% d) 1.57%
Ans: b
Solution: (40 × 10 20 × 30/ 18 × 60 × 60) × 100%
(400 600/ 18 × 60 × 60) × 100%
= (1000/ 18 × 60 × 60) × 100
= 1.54%
2. If the numerator of a fraction is increased by 20% and the denominator is decreased by 5%, the value of the new fraction becomes 5/2. The original fraction is?
a) 49/95 b) 95/48 c) 17/24 d) 24/17
Ans: b
Solution: Let the fraction = a/b
a × 120/ b × 95 = 5/2
a/b = 5 × 95/ 2 × 120
a/b = 95/48
3. There are some coins and rings of either gold or silver in a box. 60% of the objects are coins, 40% of the rings are of gold and 30% of the coins are of silver. What is the percentage of gold articles?
a) 52 b) 56 c) 58 d) 62
Ans: c
Solution :
Let total number of article = 100
coins rings
60 40
30% 70% 40% 60%
silver gold gold silver
Gold articles = 60 × 70% 40 × 40%
= 42% 16%
= 58%
4. Initially a shopkeeper had n chocolates. A customer bought 10% chocolates from n, then another customer bought 25% of the remaining chocolates. After that, one more customer purchased 20% of the remaining chocolates. Finally, the shopkeeper is left with 540 chocolates in his shop. How many chocolates were there initially in his shop?
a) 750 b) 1000 c) 1250 d) 1275
Ans: b
Solution: 10% = 10/100 = 1/10, 25% = 25/100 = 1/4, 20% = 20/100 = 1/5
10 : 9
4 : 3
5 : 4
50 : 27
27 –>; 540
1 –>; 20
50 × 20 = 1000
5. A spider climbed 62 1/2% of the height of the pole in one hour and in the next hour it covered 12 1/2% of the remaining height. If pole’s height is 320m, then the distance climbed in second hour is?
a) 10m b) 12m c) 15m d) 18m
Ans: c
Solution: 62 1/2% = 125/2 1/100 = 5/8,
12 1/2% = 25/2 × 1/100 = 1/8
Let total height of the pole be 64 units
64 –>; 3/8 24 —>; 1/8 3
total height remaining height in 2nd hour distance covered after 1hr
64 –>; 320
1 –>; 5
3 –>; 3 × 5
= 15m
6. The tank-full of petrol in Ajith’s motor cycle lasts for 10 days. If he starts using 25% more every day, how many days will the tank-full of petrol last?
a) 7 b) 8 c) 9 d) 6
Ans: b
Solution : Let us assume that Ajith uses x units of petrol everyday.
So, the amount of petrol in tank when it is full will be 10x.
x 25% of x = 125% of x = 5x/4
5x/4 5x/4 …………….. 5x/4 (n days) = 10x
n × 5x/4 = 10x
n = 8 days
To be continued
M.Venkat
Director
MVK Publications
Dilsukhnagar
7671002120