Mixed Quantitative Aptitude Questions Set 103

Directions(1-5): What will come in place of question mark “?” in the following questions.

  1. (12.16)^5.9 / (5.3)^3.11 * (2.3)^2.98 = (2.3)^(? – 3.01)
    12
    15
    10
    16
    18
    Option E
    (12.16)^5.9 / (5.3)^3.11 * (2.3)^2.98 = (2.3)^(? – 3.01)
    => (12.16)^6 / (5.3)^3 * (2.3)^3 = (2.3)^(?-3)
    => (2.3)^18 /(2.3)^6 *(2.3)^3 = (2.3)^(?-3)
    => (2.3)^15 = (2.3)^(?-3)
    => ? = 18

     

  2. 56*81 + 84*29+ ? *21 = 63*79+ 49*57
    25
    38
    22
    30
    27
    Option B
    56*81 + 84*29+ ? *21 = 63*79+ 49*57
    => 4536 +2436 + ? *21 = 4977 + 2793
    => 6972 + ? *21 = 7770
    => ? = 38

     

  3. 28% of 6525 + 35% of 6780 + 20% of ? = 52% of 5475 + 25% of 7248
    2222
    2200
    2225
    2295
    2020
    Option D
    28% of 6525 + 35% of 6780 + 20% of ? = 52% of 5475 + 25% of 7248
    => 1857 + 2373 + 0.20*? = 2847 + 1812
    => 4200 + 0.20 *? = 4659
    => ? = 2295

     

  4. (1849)^1/2 + 1155.905 / 16.927 = ?^2 + 64.89% of 95.49
    6
    4
    5
    8
    7
    Option E
    (1849)^1/2 + 1155.905 / 16.927 = ?^2 + 64.89% of 95.49
    => 43 + 68 = ?^2 + 62
    => ? = 7

     

  5. 19.8% of 780.12 + ? + 29.72% of 499.23 = 34.93% of 1359.69
    150
    170
    190
    160
    180
    Option B
    19.8% of 780.12 + ? + 29.72% of 499.23 = 34.93% of 1359.69
    => 20% of 780 + ? + 30% of 500 = 35% of 1360
    => 156 + ? + 150 = 476
    => ? = 170

     

  6. Area of the rectangular field is half the area of the circular field. Perimeter of the rectangular field is 108 meter. Find the circumference of the circular field if length and the breadth of the rectangular field are in the ratio of 11:7.
    128
    120
    130
    132
    135
    Option D
    Let the length and breadth of the rectangular field be 11x and 7x respectively.
    Now, 2(11x+7x) = 108
    => x = 3
    So, the length and breadth of the rectangular field are 33 m and 21 m resp.
    Area of the rectangular field = 33*21 = 693 m^2
    Area of the circular field = (22/7)*r^2 = 2*693
    => r = 21
    So, the circumference of the circular field = 2*(22/7)*21 = 132 m

     

  7. A shopkeeper bought an article for Rs. 1200. He marked it at somewhat above the cost price and sold it after two consecutive discounts of 25% and 20%. By what percentage he marked the article above the cost price if in this transaction he had a loss of Rs. 84?
    60%
    57%
    55%
    50%
    48%
    Option C
    Let the shopkeeper marked the article x% above the cost price.
    MP of the article = 1200*{(100+x)/100} = Rs. 1200+12x
    SP of the article = (1200+12x)*0.75*0.80 = 1200 – 84 = 1116
    720+7.2x = 1116
    => x = 55%
    The shopkeeper marked the article at 55% above the cost price.

     

  8. Abhi invested a certain amount of money in a scheme offering 20% compound interest for two years. Sahil invested Rs.1000 more than Abhi in another scheme offering 25% simple interest for two years. Find the amount invested by Abhi if the difference in the interest earned by Abhi and Sahil is Rs. 1310.
    Rs. 14000
    Rs. 11200
    Rs. 12250
    Rs. 13500
    Rs. 13000
    Option D
    Let the amount invested by Abhi and Sahil be Rs. x and Rs. (x+1000) resp.
    Interest earned by Abhi = x*{(1+0.20)^2 – 1} = Rs. 0.44x
    Interest earned by Sahil = (x+1000)*0.25*2 = Rs. 0.5x + 500
    Now, 0.5x + 500 – 0.44x = 1310
    => x = 13500
    So, the amount invested by Abhi = Rs. 13500

     

  9. Monthly incomes of A and B are in the ratio of 8:5 resp. Monthly expenditure of B is 70% more than the monthly savings of A. Monthly expenditure of A is Rs. 7800 more than the monthly expenditure of B. Find the average monthly income of A and B if the monthly savings of A and B are in the ratio of 5:4 resp.
    Rs. 18000
    Rs. 17500
    Rs. 18000
    Rs. 20000
    Rs. 19500
    Option E
    Let monthly savings be 8x and 5x of A and B resp.
    Monthly expenditure of B = 1.7*5x = Rs. 8.5x
    Monthly expenditure of A = Rs.8.5x + 7800
    Now, (5x+8.5x+7800)/(4x+8.5x) = 8/5
    => x = 1200
    Monthly income of A = 5*1200+8.5*1200 + 7800 = Rs. 24000
    Monthly income of B = 4*1200+8.5*1200+10200 = Rs. 15000
    Average monthly income of A and B = (24000+15000)/2 = Rs. 19500

     

  10. Three years ago, the average age of the family of four persons was 25 years. In between these three years, the family adopted a child. At present the average age of the family is same as that it was three years ago. Find the present age of the adopted child.
    12 years
    10 years
    14 years
    13 years
    11 years
    Option D
    Sum of the ages of the family 3 years ago = 25*4 = 100 years
    Sum of the ages of the family at present = 100 + 3*4 = 112 years
    Sum of the ages of the family at present including adopted child = 25*5 = 125 years
    Present age of the adopted child = 125 – 112 = 13 years

     


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