Number Series:

  1. The sixth number in a sequence of numbers is 32 and each number after the first number in the sequence is 4 less than the number immediately preceding it. What is the third number in the sequence?44
    1. Solution:Each number in the sequence is 4 less than the number preceding it.
    2. The equation for the terms in an arithmetic sequence is a n = a + d(n-1), where d is the difference.
    3. The sequence of six numbers – 52,48,44,40,36,32. so the third term is 44.
  2. 152 is divided into four parts is proportional to 3, 4, 5 and 7, then the smallest part is
    1. 24
      1. Let say  3x,4x,5x,7x are the four parts of the 152
      2. we can write like this 3x+4x+5x+7x=152
      3. 19x=152 , x=8
      4. So, four parts is :3×8,4×8,5×8,7×8=24,32,40,56
      5. So, the smallest part is 24.
  3. If the number 345xy is divisible by 90, then (x+y) =?
    1. 6 
    2. Factors of Divisibility of 90 = 9 and 10
    3. By the rule of divisiblity, by 9, 3+x+y=9, so x+y=6
  4. For all integers >1,<y,>=2y+(2y-1)+(2y-2)+——+1
    1. What is the value of<3> x <2>, Where X is a multiplication operator?
    2. 210
    3. Solution:
      1. <y>=1+2+3+⋯+(2y−2)+(2y−1)+2y
      2. It is arithmetic progression whose first term is a=1 and difference d=1 and last terms l=2y
      3. Sum of the AP series S=n/2(a+l)
      4. Now, find the sum  <y>=2y/2(1+2y)
      5. <y>=y(1+2y)
      6. <y>=y+2y^2→(1)
      7. Now find the value of  <3>=3+18=21 and <2>=2+8=10
      8. The value of <3>×<2>=21×10=210 (Or)
      9. according to the above function for <2> = 4+3+2+1= 10 & for <3> = 6+5+4+3+2+1=21. so the final value is 21*10 = 210.
  5. If  0.75x = 0.02y, then the value of ( y-x/y+x) is
    1. 73/77
    2. Solution:
      1. Given, 0.75x=0.02y
      2. y=0.75x/0.02, y = 75x/2—1
      3. Sub 1 in y-x/y+x = 37.5x-x/37.5x+x = 36.5*2/38.5*2 = 73/77
  6. The product of two numbers a and b is twice the sum of the numbers. What is the sum of the reciprocals of a and b?
    1. 1/2
    2. Solution:
      1. From given question we have,ab=2(a+b)
      2. 1/a+1/b = a+b/ab = a+b/2(a+b) = 1/2
      3. Hence sum of reciprocals of a and b is 1/2
  7. If E=5  and HOTEL=12, how will you code BALM?
    1. 7
      1. E = letters position/Number of letter = 5/5 = 1
      2. HOTEL = 8+15+20+5+12/5 = 60/5 = 12
      3. BALM = 2+1+12+13 / 4 = 28/4 = 7

Ratio:

  1. What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 9 years and that for 12 years?
    1. 3:4
    2. Solution:
      1. Let the principal be P and rate of interest be R%.
      1. SI for 9 years = P*9*R/100
      2. SI for 12 yrs = P*12*R/100, so 3:4
  2. On a map, 1 inch represents 250 miles. What is the actual distance between two cities if they are 4.5 inches apart on a map?
    1. 1125 miles
    2. 1 inch = 250 miles
      1. so 9/2 inches = 250*9/2 = 1125 miles
  3. If x and y are positive numbers and x is 25 % greater than y, what is the value of the ratio y/x?
    1. 0.80
    2. Solution:
      1. Let the number y be=100
      2. x=100+25 =125
      3. Now, y/x=100/125 = 4/5 = 0.80
  4. A group of 40 employees of Cadre A has a mean age of 31. A different group of 60 employees of Cadre B has a mean age of 28 years.What is the mean age of two employees of the two groups together?
    1. 29.2
    2. Solution:
      1. Ratio of number of persons in both groups = 40:60 = 2:3
      2. the means age of  employees of 2 groups  = 2*31 + 3*28 / 5 = 146/5 = 29.2 years.

Propositions and Percentage:

  1. In a bag, there are coins of 5ps, 10ps, and 25ps in a ratio of 3:2:1. If there are Rs.60 in all, how many 5ps coins are there?
    1. 300
    1. Solution:Reference:
      1. Let the number of coins of 25, 10 & 5 Paise be p : 2p : 3p.
      2. 60Rs = 6000 paise
      3. 25p : 10p : 5p = 1 : 2 : 3
      4. 25*X + 10*2X + 5*3X = 6000
      5. 60x = 6000, x = 100
      6. So number of 5 paise coins = 3*100 = 300 coins.
  2. During one year, the population of a town increased by 5% and during the next year, the population decreased by 5%. If the total population is 29925 at the end of the second year, and then what was the population size at the beginning of the first year?
    1. 30,000
    2. Solution:
      1. Let the population at beginning of first year be x.
      2. At the end of 1st year, the population was 105/100*x
      3. At the end of 2nd year, the population was = 1.05x * 0.95 = 21/20*x*19/20 = 399/400 x
      4. 399/400 x = 29925 So, x=30,000
  3. 5 % of the inhabitants of a village had died of malaria, panic set in. During this 20% of the remaining inhabitants left the village. The population is then reduced to 4750. Find the number of original inhabitants
    1. 6250
    2. Let the total number of original inhabitants be x
    1. ((95/100) * (80/100)*x) = 4750
    2. Number of original inhabitants = 19/25 *x = 4750 = x= 118750/19 = 6250
  4. In a class, 20% of the boys have blue eyes and 10% of the girl has blue eyes. If the ratio of boys to girls in the class is 4:3, then what is the portion of the students in the class having blue eyes?
    1. 11/70
      1. Solution:
      2. Number of  all boys who are blue-eyed => 1/5 * 4/7 = 4/35
      3. How many of all girls who are blue-eyed => 1/10 * 3/7 = 3/70
      4. Total students who are blue-eyed = 4/35 + 3/70 = 11/70
  5. A trader mixes three varieties of refined oil costing Rs. 100, Rs. 40 and Rs. 60 per liter in the ratio 2:4:3 of a liter, and the sells the mixture at Rs. 66 per liter. What percentage of profit does he make?
    1. 10%
    2. Solution:
      1. Given that, the Ratio of varieties of refined oil is 2:4:3 costs are Rs. 100, Rs. 40, Rs. 60 per liter
      2. So, Total cost would be
      3. 2x*100+40*4x+60*3x =200x+160x+180x = 540x
      4. Given – Sells the mixture at Rs. 66 per liter
      5. So, the total selling price becomes (2x+4x+3x)*66 = 9x*66 = 594x
      6. So, Selling price is more than cost price.
      7. So, Profit percentage would be 594x-540x/540x * 100=54/540 *100=10%
  6. A retailer marks all his goods at 40% above the cost price and thinking that he will still make a 20% profit, offers a discount of 20% on the marked price. What is his actual profit percentage on the sales?
    1. 12%
    1. Solution:
      1. Let C.P. = Rs. 100.  Then,marked price  = Rs. 140.
      2. S.P. = 80% of Rs. 140 = Rs. 112
      3. Gain % =  12%
  7. Manoj’s commission is 10% on all sales up to Rs. 10,000 and 5% on all sales exceeding this. He remits Rs. 75,500 to his company after deducting his commission. The total sales will come out to be
    1. Let, the total sales of the company be x
    2. Given: Amount Manoj has remitted = Rs. 75500
    3. Now, commission for 10,000 = 10,000 × (10/100) = Rs. 1000
    4. Rest amount = (x – 10000)
    5. Commission on rest amount = (x – 10000) × 5/100
    6. So, x – [1000 + (x – 10000) × 5/100] = 75500
    7. ⇒ x – x/20 – 500 = 75500 ⇒ 19x/20 = 76000 ⇒ x  = (76000 × 20)/19 = 80000
    8. Therefore, the total sales is Rs. 80,000.
  8. 40% of the employees in a factory are workers. All the missing employees are executives. The annual income of each worker is Rs. 39,000. The annual income of each executive is Rs. 42,000. What is the average income of all employees in the factory?
    1. Rs 40,800
    2. Solution: Let the factory have X employees.
    1. We are given that 40% of the employees are workers.
    2. The number of worker employees = 40/100 = 2x/5
    3. Remaining number of executive employees = 
      1. (Total number of Employees) − (The number of Workers)
      2. X − 2X/5 = 3X/5
    4. The annual income of each worker is Rs. 390
    5. Hence, the total annual income of worker employees = 2X/5 × 390 = 156X
    6. The annual income of each executive is Rs.420.
    7. Hence, the total income of all the executives together is, = 3X/5 × 420 = 252X
    8. Hence, the total income of the employees is, = 156X + 252X = 408X
      1. Method-2:
      1. On considering the factory have 100 employees,
      2. Avg annual income= sum of income of all employees/Number of employees
      3. 39000*40 + 42000*60/60+40  = (15600+25200)  = Rs.40,800

Interest and Discounting:

  1. The simple interest on a certain sum of money for 3 years at 4% per annum is half the compound interest on Rs. 2,000 for 2 years at 10% per annum. The sum invested in simple interest is
    1. Rs 1,750
    2. Solution:
      1. A= Rs.[2000×(1+10/100)2] = rs. 2420
      2. CI = A-P = 2420 -2000 =Rs.420 Since half of CI is SI
      3. As given, SI = 420/2 = rs. 210
      4. By formula = SI = PRT/100
      5. ∴Sum=Rs.(210×100/3×4) =Rs.1750
  2. If the simple interest on a certain sum for 1 year 3 months at 8 1⁄2% per annum exceeds the simple interest on the same sum for 8 months at 7 1⁄2 % per annum by Rs. 45, then the sum is
    1. Rs.800
    2. (x*1 1/4 *8 1/2) / 100 – (x 2/3 * 15/2) / 100 = 45
    1. x*5/4*17/2-5x = 4500
    2. 35x/8 – 5x = 4500
    3. 85x – 40x / 8 = 4500
    4. solve x = 800
  3. The simple interest on a certain sum of money at the rate of 5% p.a. for four years is Rs. 420. At what rate of interest, the same amount of interest can be received on the same sum after two and a half years?
    1. 8%
    2. Solution:
    3. Given, S.I. = Rs.420, R = 5% T = 4 years
      1. Principal = (100 *420/4*5) = Rs.2100
      2. Rate = 420 *100*2/2100*5 = 168/21 = 8%
  4. The sum of money doubles at compound interest in 6 years. In how many years will it become 16 times?
    1. 24 years
    2. Solution
      1. Let the amount of money be Rs.P and the rate of interest be R% p.a. A=P(1+r/100)^n
      2. The amount at the end of 6 years = 2P
        1. P(1+r/100)^6=2P
        2. (1+r/100)^6 = 2 —-1
        3. Money becomes 16 times after n years
        4. [(1+r/100)^6]^4 = (2^4)
        5. (1+r/100)^24 = 16
        6. So n= 24 years, After 24 yrs money become 16 times
  5. In a new budget, the price of petrol rose by 25%. how much percent must a person reduce his consumption so that his expenditure on it does not increase?
    1. 20%
    1. Solution:
      1. Formula Used:
      2. Expenditure = Price × Consumption
    2. Calculation:
      1. We know that,
      2. Fraction value of 25% = 1/4
      3. Let the original price be 4x 
      4. Price is increased by 25%
      5. New price = 4x + (25/100) × 4x = 5x 
      6. Ratio of original price to new price = 4 : 5
      7. As the expenditure does not change so price will be inversely proportional to consumption
      8. Ratio of old consumption to new consumption = 5 : 4 
      9. Decrease in consumption = 5 – 4 = 1 units 
        1. ∴ Percentage decrease in consumption = (1/5) × 100 = 20%
    3. Method:2
    1. Let Original price of 1 ltr petrol = rs.100
      1. New price = rs.125
        1. Qty of petrol in rs.100 = 1/125*100 = 4/5 ltrs.
        2. % deduction in consumption,
          1. 1-4/5/1 * 100 = 100/5 = 20%
  6. At what percentage above the cost price must an article be marked so as to gain 8% after allowing a customer a discount of 10%?
    1. 20%,
    2. Solution:
      1. Let CP of an article is rs.100 to get a profit of 8%
      2. SP = rs.108
      3. Now, discount = 10%
    1. M.P = SP*100/100-Dis%
    2. =108*100/100-10 = 108*100/90 = rs.120
    3. Reqd % = 120-100/100 * 100 = 20%
    4. Note:
      1. If M.P is given = 100 – discounted %/100 * M.P
      2. If S.P is given = 100/100-Discounted% * S.P
  7. Sanjay sold an article at a loss of 25%. If the selling price had been increased by Rs. 175, there would have been a gain of 10%. What was the cost price of the article?
    1. Rs. 500
    2. Selling price of the article = {100-L}/{100} * C.P
    1. = {100-25}/{100} * x
    2. = {75}/{100}x= {3x}/{4} …… (i)
    3. It is also given that if the S.P. of the article is increased by Rs. 175 and then it is sold then there is a gain of 10%, therefore,
    4. S.P. = 100+G/100*C.P
    5. 3x/4  + 175 =  110/100 *x [substituting G = 10%  and value of S.P. from (i)]
    6. ⇒ 0.75x + 175 = 1.1x
    7. 0.35x = 175
    8. x = Rs. 500
      1. Method-2:
      2. (10+25)% increase of CP = 175
      3. C.P = 175*100/35 = rs.500
  8. A person incurs a 10% loss by selling a watch for Rs.1,800. At what price should the watch be sold to earn a 10% profit?
    1. Rs. 2200
    2. Solution:
      1. S.P = 1800, Loss = 10%
      2. C.P = S.P*100/100-loss = 1800*100/90% = 2000
      3. to get a profit of 10%
      4. S.P = C.P*(100+profit%/100) = 2000*110/100 = Rs.2200

Average:

  1. If the average of five numbers is -20 and the sum of three of the numbers is 32, then what is the average of the other two numbers?
    1. -66,
    2. Solution:
      1. Let the three numbers be a,b,c.
      2. Average of 5 numbers = -20
      3. Total sum of 5 numbers = -20*5 = -100
      4. Sum of 3 numbers,  = 32, let it be a+b+c = 32
      5. Sum of other two numbers = -100-32 = -132
      6. Average of other 2 numbers = -132/2 = -66
  2. An airplane flies along the four sides of a square field at the speed of 400,600,800 and 200 km/hr. find the average speed of the plane around the field.
    1. 384 km/hr
    2. Solution:
      1. average speed = total distance / total time
      2. x+x+x+x / x/400 + x/600 + x/800 + x/200
      3. 4x/(25x/2400)
      4. 4x * 2400/25x = 4*96 = 384 km/hr
      5. Method-2:
        1. Let the speed of aeroplane is taken as 400, 600, 800 and 200 km/hr
        2. All the sides of  a square are same so equal distance at different speed
        3. Time taken for the first side = 2400/400 = 6 hrs
        4. For the second side = 2400/600 = 4 hrs
        5. For the third side, 2400/800 = 3 hrs
        6. For the fourth side = 2400/200 = 12 hrs
        7. Total distance = 4*2400 = 9600 kms
        8. Total time taken = 6+4+3+12 = 25 hrs
        9. Avg speed = Total distance/ Total time = 9600/25 = 384 kms/hr
  3. The average age of a class of 59 students is 18 years. 1f the age of the teacher is included, and then the average increases by 3 months. The age of the teacher is:
    1. 33 years
    2. Solution: Since the average age of a class of 59 students is 18 yrs, 59*18 = 1062 months
    1. When the teacher’s age is included the avg age increases by 3 months, 18.25*60 = 1095
    2. the age of the teacher = 1095-1062 = 33 yrs
      1. Method-2:
      2. Avg age of students = 18 yrs
      3. Total after including teacher= 59+1 =60
      4. Total increase of 3 months = 60*1/4 = 15 yrs
      5. This increase of 15 yr due to the inclusion of a teacher
      6. Thus the age of a teacher = 18+15 = 33 yrs

Measurements:

  1. A man traveled from a village to a railway station at the rate of 20Kmph and walked back at the rate of 5Kmph. If the whole journey took 5 hours, find the distance of the railway station from the village.
    1. 20 Km
    2. Solution:
      1. Let the distance From village to railway station be x
      2. x/20 + x/5 = 5
      3. x+4x/20 = 5, Hence 5x = 100, So x = 20 Km
  2. Sameer walks a certain distance and rides back taking a total time of 40 minutes. He could walk both ways in 60 minutes. How long would it take to ride both ways?
    1. 20 min
    2. Solution:
      1. Walk + ride back = 40 minutes.     —————- (1)
      2. Walk + Walk = 60 minutes.
      3. 2 * walk = 60 minutes. So for walking, it takes 30 minutes
      4. Time taken to ride both ways, Riding = 40-30 = 10 *2 = 20 mins
  3. Rohan gave three-fifth of the amount he had to Sahil. Rohan has now 2000 rupees. How much did he give to Sahil?
    1. 3000 rupees
    2. Solution:
      1. Let Rohan had initially rs.x
      2. Amount Rohan gave to Sahil = 3/5x
      3. Remaining amount x-3/5x = 2000
      4. 2x/5 = 2000, x = 2000*5/2 = rs.5000
      5. Since the Amount given to Sahil = 3/5 *x
      6. 3/5 *5000 = rs.3000
  4. A train A leaves Agra at 5 a.m. and reaches Delhi at 9 a.m. Another train B leaves Delhi at 7 a.m. and reaches Agra at 10:30 a.m. At what time do the two trains cross each other?
    1. (1) 6:36 a.m.
    2. (2) 6:56 a.m.
    3. (3) 7 a.m.
    4. (4) 7:56 a.m.
      1. Solution:
      2. Suppose the distance between Agra to delhi is x km
      3. The speed of 1st train which moves from Agra to delhi = x/4 km/hr
      1. Speed of second train coming from delhi to agra = x/3 1/2 = 2x/7 km/hr
      2. Relative speed of both trains = 2x/7+x/4 = 15x/28
      3. According to qtn,
      4. second train runs 2 hrs late relative to the first train, thus when the second train started, first train would have run half the dis betwen delhi and agra
      5. Now total speed after 2 hrs for travelling a dis x = x/2 km
      6. So time taken = x/2 / 15x/28 = 14/15 hr, 14/15*60 = 56 min
      7. Thus the time of meeting of both trains = 7:56 A.m
      8. Method:2
        1. Let the distance between Delhi and agra be x km and let the trains meet y hrs after 7 a.m
        2. So, A covers x km in 4 hrs and B covers x km in 7/2 hrs
        3. Speed of A = x/4km/hr, Spped of B = 2x/7 km/hr
        4. Dis covered by A in y+2 hrs+ Dis covered in y hrs = x
        5. x/4(y+2) + 2x/7*y = x
        6. y+2/4+2y/7=1
        7. y=14/15 hr = 14/15*60 = 56 min
        8. Hence the train meet = 7hrs56 min
  5. Which of the following fraction is the result of the sum of an integer and its reciprocal?
    1. 15/8
    2. 26/5
    3. 36/7
    4. 37/5
      1. 15/8 = 8x^2+8 = 15x so multiplying 64 and -15
      2. 26/5 = 5x^2-26x+5 = 0, 5x^2-25x-x+5 = 0 5x(x-5) – (x-5), (5x-1)
        1. (x-5) so x=1/5 or x=5
      3. 36/7 = 7x^2+7 = 36x, so multiply gives 49 and addtion -36
      4. 37/5 = 5x^2+5 = 37x, 
  6. Two cyclists start from the same place in opposite directions. One goes towards the north at 9 kmph and the other goes towards south at 10 kmph. What time will they take to be 47.5 km apart?
    1. 2 1⁄2 hrs          
      1. Solution:
    1. Relative speed for a cyclist moving in opposite direction = 9+10 = 19 km/hr
    2. So, both cycles are at a distance of 19 km in one hr
    3. Time taken to be 47.5 km apart, 47.5/19 = 2hrs 30 min
  7. In a class of 35 students, Rajan is placed seventh from the bottom whereas Karan is placed ninth from the top. Divya is placed exactly in between the two. What are the differences in the positions of Rajan and Divya?
    1. 9
    2. 10
    3. 11
    4. 13
    1. Solution:
      1. Rajan’s rank from top  = 35-7+1 = 29th Divya’s placed exactly between karan and rajan
      2. exactly middle places  = 29-9-1 = 19th place so, divya’s rank is 19th from top.
      3. Now, difference in rajan’s rnak and divya’s rank = 29-19 = 10
  8. The price of item A increases by 50 paise every year while the price of item B increases by 25 paise every year. If in 2008, the price of item A was Rs. 3.20 and that of B was Rs. 5.30, in which year item A will cost 40 paise more than item B?
    1. 2018
      1. Solution:
    1. Price of item A after n years,
      1. 3.2+0.5n
    2. Price of item B after n years,
      1. 5.3+0.25n
    3. According to question = (3.2+0.5n) – (5.3+0.25n) = 0.4
      1. n=2.5/0.25 = 10 years
      2. So, 10 years after 2008 is 2018
    4. Reference:
UGC NET previous year questions
UGC NTA NET Paper-1