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DIRECTION: What should come in place of the question mark (?) in the following question?,
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DIRECTIONS: In the given question two expressions numbered I and II are given. You have to solve both the expressions and give answer: ,
I. 1596 × 196 ÷ 49 – 484 ,
II. 81 × 49 + (4)3 × 29
DIRECTION: What should come in place of the question mark (?) in the following question? ,
(5 x 7)% of (34 x 55) + 456.60 = 699.1 + ?
DIRECTIONS: In the given question two expressions numbered I and II are given. You have to solve both the expressions and give answer:
I. (243)2 ÷ (27)2 × 32 ÷ 12
DIRECTIONS: Find the missing term in the following series. ,
11240, 9524, 8204, 7214, ?, 5990
11240 – (1716) = 11240 – (1728 – 12) = 9524
9524 – (1320) = 9524 – (1331 – 11) = 8204
8204 – (990) = 8204 – (1000 – 10) = 7214
7214 – (720) = 7214 – (729 – 9) = 6494
6494 – (504) – (512 – 8) = 5990
5111808, 159744, 9984, 1248, ?, 156
5111808 ÷ 32 = 159744
159744 ÷ 16 = 9984
9984 ÷ 8 = 1248
1248 ÷ 4 = 312
312 ÷ 2 = 156
363, ?, 431, 516, 635, 788
363 + 17 × 1 = 380
380 + 17 × 3 = 431
431 + 17 × 5 = 516
516 + 17 × 7 = 635
635 + 17 × 9 = 788
856, 874, 899, ?, 990, 1066
856 + 18 = 874
874 + (18 + 7) = 874 + 25 = 899
899 + (25 + 7 + 5) = 899 + (25 + 12) = 899 + 37 = 936
936 + (37 + 12 + 5) = 936 + (37 + 17) = 936 + 54 = 990
990 + (54 + 17 + 5) = 990 + (54 + 22) = 990 + 76 = 1066
1066 + (76 + 22 + 5) = 1066 + (76 + 27) = 1066 + 103 = 1169
122, ?, 741, 2968, 14845, 89076
122 × 2 + 2 = 246
246 × 3 + 3 = 741
741 × 4 + 4 = 2968
2968 × 5 + 5 = 14845
14845 × 6 + 6 = 89076
A bag contains a total of 70 coins, in denominations of 50 paise coins and Rs.2 coins, totally worth Rs. 80. Find the number of 50 paise coins in the bag.
Let the number of 50p coins and Rs.2 (200p) coins be a and b.
Total value = Rs. 80 = 8000p
Therefore, 50a + 200b = 8000
a + 4b = 160---------(1)
And a + b = 70--------(2)
Subtract (2) from (1)
3b = 90
b = 30 and a = 40
If the height of a cyllinder becomes 6 times and area of base becomes 1/9 times, then curved surface area of the cyllinder will become how many times?
Two times Abhi’s age is thrice the age of Balu and four times Balu’s age is five times the age of Srikant. Also six times Srikant’s age is four times the age of David. If Abhi’s age is 30 years, what is David’s age?
Let the ages of Abhi, Balu, Srikant and David are a, b, s and d respectively.
Given, a = 30 years.
2a = 3b; b = (2/3)a = (2/3) × 30 = 20 years.
4b = 5s; s = (4/5)b = (4/5) × 20 = 16 years.
6s = 4d; d = (6/4)s = (6/4) × 16 = 24 years.
There are 50 balls numbered 1 to 50 in a tray. Every ball bearing a number which is a multiple of some other number ‘n’ (n > 1) is removed and put in a basket. What is the total number of balls removed?
The balls having prime number are not removed.
Also the ball with number 1 is not removed
Number of prime numbers between 1 to 50 = 15
Number of balls that are removed = 50 – (15 + 1) = 34
Pawan and Raman started a business by investing Rs.18000 and Rs.25200 respectively. What is the share of Pawan in the profit of Rs.8640 earned at the end of the year?
The ratio of the shares of Pawan and Raman in the profit = Rs.18000 : Rs.25200 = 5 : 7
Therefore, share of Pawan in the profit = (5/12) x Rs.8640 = Rs.3600
Anil marks up the price of an article by 50% and then offers a discount of 20% to Raman. Raman sells it for Rs.20 more than the price at which he purchased it. If Raman’s selling price is 30% more than the original cost price of the article, then Raman’s profit percentage is
Let the cost price of Anil = Rs.a
Then his marked price = Rs.1.5a
and his selling price = Rs.1.5a(0.8) = Rs.1.2a
Raman’s cost price = Rs. 1.2a
Raman’s selling price = Rs.(1.2a + 20) = Rs.1.3a
⇒ a = 200
∴ Raman’s cost price = 1.2a = Rs. 1.2(200) = 240
∴ Raman’s profit % = (20/240) x 100% = 8.33%
A train X running at a speed of 60 kmph crossed another train Y of length 540 m running in the opposite direction at a speed of 72 kmph in 33 seconds. What is the length of train X?
let the length of train X = ‘a’ m
Relative speed of train X with respect to train Y = (60 + 72) x 5/18 m/s = 110/3 m/s
Therefore, (a + 540) = (110/3) x 33
a + 540 = 1210
a = 1210 – 540 = 670 m
Variety A of tea costs Rs. 180 per kg. Variety B of tea costs Rs. 270 per kg. Variety C of tea was formed by mixing A and B in the ratio 3 : 2. Variety D of tea was formed by mixing A and B in the ratio 5 : 4. C and D were mixed in a certain ratio and the resulting mixture was sold at Rs. 261 per kg at 20% profit. Find the ratio in which C and D were mixed.
Costs of tea of variety of C = = Rs.216/kg
Costs of tea of variety of D = = Rs.220/kg
S.P. of mixture of C and D = Rs.261/kg
And profit = 20%
C.P. of mixture of C and D = = Rs.217.5/kg
Let the required ratio = x : y, then
= 217.5
216x + 220y = 217.5x + 217.5y
2.5y = 1.5x
x : y = 5 : 3
In a class of 300 students, 130 play football, 160 play hockey, 150 play cricket, 70 play hockey and cricket, 40 play football and cricket, 84 play football and hockey and 16 play all the three games. ,
The number of students who do not play any of the three games is
Number of students who play = 22 + 22 + 56 + 68 + 24 + 54+ 16 = 262
Number of students who do not play any game = 300 – 262 = 38
The ratio of copper and zinc in a 63 kg alloy is 4 : 3. Some amount of copper is extracted from the alloy and the ratio becomes 10 : 9. How much copper is extracted.
Copper: Zinc
4:3
Copper = (4/7) × 63 = 36 kg
Zinc = (3/7) × 63 = 27 kg
Let x kg copper is extracted
Remaining copper = 36 – x kg
New ratio = (10/9)
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