Number Series:

  1. The number X in the following series:
    1. 3,8,18, 38,X,…is
      1. 78, 80, 76, 74
        1. 3×2=6+2=8
        2. 8×2=16+2=18
        3. 18×2 =36+2=38
        4. 38×2=76+2=78
  2. Missing term in the given series 2,4,6,12,10,36,14,?,18,324
    1. 108
    2. The pattern of 1st group – is 2+4 = 6+4 = 10+4 = 14+4= 18 so, 2,6,10,14,18
    3. 2nd group= 4,12,36,?,324 = 4*3 = 12*3=36*3 = 108*3 = 324
  3. If HPU = 9, JNU=9, then DU = ?
    1. 5
      1. HPU = 8+16+21 = 45 / 5 = 9
      2. JNU = 10+14+21 = 45/5 =9
      3. Letters are divided by 5 by adding the alphabetical order.
  4. Which one is the wrong number in the given series?
    1. 3,13,43,53,63,83
      1. 63
      2. Except 63 all are prime numbers
  5. The sum of a number and its inverse is -4. The sum of their cubes is -52.
  6. The following number would occur in the series 2,6,30,60,130,210,—–?
    1. Options: 330, 350, 370, 378
    1. Solution:
      1. 1^3+1=2
      2. 2^3 -2 =6
      3. 3^3 +3=30
      4. 4^3 -4=60
      5. 5^3+5=130
      6. 6^3-6=210
      7. 7^3+7 = 350
  7. The smallest integer is greater than 1 which is simultaneously a square and a cube of certain integers is 64.
  8. Determine the term ‘x’ in the following series – 2,9,20, 35,54,77,x – 104
    1. 104
    2. Solution:
      1. 2 + 7 = 9
      2. 9 + 11 = 20
      3. 20 + 15 = 35
      4. 11 is 4 more than 7 and 15 is 4 more than 11.
    1. If this is the pattern, each new number should be increased by a number that is 4 more than the last increase and the next numbers would be:
      1. 35 + 19 = 54
      2. 54 + 23 = 77
      3. 77 + 27 = 104 and the series is 2, 9, 20, 35, 54, 77, 104.
  9. Identify the term X in the following series,
    1. 3,2, 11, 14, 27, 34, 51, X
    2. Options:
      1. 62
      2. Solution:
      3. 1^2+2, 2^2-2, 3^2+2, 4^2-2, 5^2+2, 6^2-2
      4. 3 = 1^2+2 = 1+2=3
      5. 2 = 2^2-2=4-2=2
      6. 11 = 3^2+2=9+2=11
      7. 14 = 4^2-2 = 16-2 = 14
      8. 27 = 5^2+2=25+2=27
      9. 34 = 6^2-2 = 36-2 =34
      10. 51= 7^2+2 = 49+2 = 51
      11. Similarly,
      12. x = 8^2 – 2= 64-2 =62

Ratio:

  1. 2 numbers are in the ratio 3:7. If 8 is added to both the numbers their ratio becomes 5:9. The numbers are 
    1. 12,28
    2. Solution: Let the two numbers be x and y
    3. x/y = 3/7, so x=3y/7—1
      1. x+8/y+8= 5/9–2
      2. From 1, x =3y/7 and from 2, 5y-9(3y/7)
      3. So y=28 and x = 12.
  2. The ratio of 2 numbers a and b is 3:7. After adding 9 to each number the ratio becomes 9:17. The numbers a and b are (18,42)
    1. 3x+9/7x+9 = 9/17
    2. 51x+153 = 63x+81
    3. By solving X = 6, Numbers A and B is 3*6 = 18 and 7*6 = 42 which means (18,42)
  3. If x:y = 7:9 then 3x-5y: 4x+y = ?
    1. -24/37
    2. If x=7k and Y=9k sub in the eqn, 21k-45k/28k+9k
    3. -24/37
  4. In what ratio must a vendor mix 2 varieties of sugar worth rs.60 /kg, and rs.78/kg, so that by selling the mixture at rs.76.8/kg the vendor gains 20%
    1. 7:2
    2. Solution:
      1. S.P. of 1 kg of the mixture = Rs.76.8, Gain = 20%
        1. C.P. of 1 kg of the mixture = Rs.100/120 * 76.8 = rs.64
      2. Given, 1st variety = rs.60/kg 2nd Variety = rs. 78/kg
      3. Mean price = 64
      4. By the rule of the allegation,
        1. 78-64 = 14 and 64-60 = 4, therefore =14/4= 7:2
  5. If income X is 20% more than that of Y and income of Y is 25% less than that of Z, then the total income of X, Y and Z are in which of the following ratios
    1. 18:15:20,
    2. Solution:
      1. let the income of z be 100
      2. y’s income = when y is 25% less than z = 75/100 *100 = rs.75
      3. x’s income = 20/100*75 = 15+75 = rs.90
      4. x:y:z = 90:75:100 = 18:15:20

Proportion and Percentage:

  1. Average of two numbers a and b is 22. 60% of a = 50% of b then what will be the product of a and b?
    1. Avg of 2 numbers a and b = 22
      1. Sum of a and b = 22*2 = 44 (a+b=44)
    2. 60% of a = 50% of b
      1. a*60/100 = b*50/100, a/b = 5/6
      2. a:b = 5:6
    3. Let us suppose a = 5k and b=6k then from eqn 1
      1. 5k+6k = 44=K=1 so a = 5*4 = 20 and b= 6*4 = 24
      2. the product of a and b (a*b) = 20*24 = 480
  2. If 3% of (a+b) = 7% of (ab) and 5% of (a-b) = 4% of ab then what % of b is a? 4700/23%
    1. Solution:
    2. 3(a+b)/100 = 7 ab/100
    3. 5(a-b)/100 = 4 ab/100
    4. 3(a+b)/5(a-b) = 7/4
    5. 3a+3b/5a-5b = 7/4
    6. 12a+12b = 35a-35b
    7. 23a = 47b so a/b = 47/23
    8. Since asked % of b is a, a/b*100 =  47 * 100/23
  3. The product of 2 consecutive integers is divisible by 2. Therefore (4*5=20)is divisible by 2. This is an example of (General to specific)
    1. Deductive reasoning
  4. In a class, there are 60% of female students and the remaining are male, 40% of the female students and 70% of the male students have opted for the research methodology course. If in all, 60 students have not opted for the research methodology course, the total number of students in the class?125
    1. Solution:
    2. Let the total number of students in the class be x
    3. So, [x-(70/100(40/100*x) + 60/100(40/100*x)] = 60
    4. x-[14x/50 + 6x/25] = 60
    5. x- [ 14x+12x/50] = 60
    6. x-26x/50 = 50x-26x/50
    7. 24x = 3000 = 125
  5. A student obtains overall 78% marks in his examination consisting of physics, chemistry, math, computer science, and English. Marks obtained in each subject and the maximum marks in the following table
SubjectPhysics (out of 200) Chem (out of 100) Math (out of 200) Comp Sci (out of 200) Eng (out of 100)
Marks obtained15580165140x
Class subjects

The marks (x) obtained by the student in gen. English would be – 84

  1. Solution:
    1. Total maximum marks  = 200+100+200+200+100 = 800
    1. Overall marks % = 800*78/100 = 624
    1. Total marks obtained = 540+x
    1. So marks in english, 540+x=624, Hence X = 624-540  = 84
  2. A shopkeeper sells a refrigerator for Rs. 22, 000.00 and makes a profit 10% If he desires to make a profit of 18% what should be his selling price?
    1. rs.23,600
    2. Solution:
    1. 110/100 *x = 22000, Hence
    2. x = 22000 * 100 / 110 = Rs. 20,000
    3. If the cost price is 20,000 for obtaining a gain of 18%
    4. S.P = 118/100 * 20,000 = Rs.23,600

Interest and Discounting:

  1. A sum of Rs.2000 yields rs.180 with a simple interest in 9 months the annual rate of interest is
    1. 12%
    2. By using formula ⇔3000×9/12×R/100=180
    3. Therefore R = 12%
  2. A sum of money with C.I becomes Rs. 2,400 in 1 year and 3,000 in 2 years. Find out the principal amount.
    1. 1920
      1. Solution:
      2. Amount (A) = p (1 + r/100)n
        1. Where p, r and n respectively are principal, rate of interest and time
      3. Calculation:
    1. Initial year it is SI
      1. 600 = 2400*1*R/100
      2. R=25%
        1. According to the question,
          1. 3000 =  P (1 + 25/100)2
        2. P = 3000 *(125/100)^2
        3. P = rs.1920
  3. A certain principal invested at C.I payable yearly amounts to rs.10816.00 in 8 yrs and rs.11248.64 in 4 years. What is the rate of interest?
    1. 4%
      WKT, A = P+C.I, So, A1= 11248.64, A2 = 10816
      1. A = P(1+r/100)^n
      2. 10816 = P(1+r/100)^3
      3. 11248.64 = P(1+r/100)^4
      4. On solving,
      5. 1124.64/10816 = (100+r/100)
      6. R= 4%
  4. The exact equivalent of 3 successive discounts of 10%, 15%, 20% by the sale of an article – 38.8%
    1. Formula:
      1. If x% discount is offered,
      2. Final price = (100 – x)/100 × Actual Price
      3. Discount = Actual price – Final price
    2. Calculation:
      1. Let the actual price of the article be 100.
      2. Price after 1st discount = 90/100 × 100 = 90
      3. Price after 2nd discount = 85/100 × 90 = 76.5
      4. Price after 3rd discount = 80/100 × 76.5 = 61.2⇒ Discount = 100 – 61.2 = 38.8
      5. ∴ Discount percent = 38.8/100 × 100 = 38.8%
  1. The % profit gained by a vendor when he sells an article?
    1. a. He gives a 10% discount on the market price
      b. He would have earned a 25% profit if he did not give any discount.
      c. The cost price of the article is rs.8000
      Choose the correct option
      1. a and b only
  2. The C.I on a certain sum of money for 3 years at the rate of 10% p.a is rs.908.00 The SI on the same at the rate of 8% p.a for 4 years would be equal to
    1. rs.960
    2. Solution:
      1. For 2nd year, 10+10 + 10*10/100 = 20.1
      2. Effective rate of SI For 3rd Year = 20.1+10 + 0.201 = 30.3
      3. The effective rate of SI for 4 years = 4*8 = 32
      4. Reqd S.I = 908/30.3 * 32 = ~959=960 Or
    3. Method-2:
    1. C.I = P(1+R/100)^n – P
      1. 993 = P[(11/10)^3 – 1]
      2. P = 993*1000/331 = rs.3000
      3. So SI=P*n*R/100 = rs.960
  3. A person buys 3 items of rs. 1500 each and 2 items of rs.1800 each form a shopping mall. On the total bill, a discount of 10% is given. What is the average cost of items?
    1. 1458
    2. Solution:
      1. The total cost of 3 items @ the rate of 1500 each = 3*1500 = rs.4500
      2. The total cost of 2 items @ the rate of 1800 each  =2 * 1800 = Rs.3600
      3. Total cost of 5 items  = Rs.4500 + 3600 =  rs.8100
      4. After 10% Discount on total, Net cost = 8100 * (100-10/100) = Rs.7290
      5. The average cost of items = 7290/5 = Rs. 1458.
  4. At what % above the C.P must an article be marked so as to gain 8% after allowing a customer discount of 10%?
    1. Let the MP be rs.100
      1. Discount on MP = 10% of 100
      2. SP = 100-10 = rs.90
      3. C.P = 100/108*90 = rs.250/3
      4. Price difference of MP and CP = 100-250/3 = 50/3
      5. % of CP (that should be kept above CP) = 50/3 / 250/3 *100 = 20%

Averages:

  1. Average of 2 numbers a and b is 22 and 60% of a = 50% of b. The product of a and b is
    1. 480
      1. Avg of 2 numbers a and b = 22
      2. Sum of a and b = 22*2 = 44
      3. 60% of a = 50% of b
      4. a*60/100 = b*50/100, a/b = 5/6 or a:b = 5:6
      5. Let us suppose a = 5k and b=6k then from eqn 1
      6. 5k+6k = 44=K=1 so a = 5*4 = 20 and b= 6*4 = 24 the product of a and b = 20*24 = 480
      7. Reference
  2. What is the average age of 30 students studying in a class?
    1. Given: B:G (Boys: Girls) = 7:8
    2. Number of boys in the class
      1. 30*7/15 = 14
      2. Number of girls = 30*8/15 = 16
    3. Given:The average age of boys and girls – 12.2 yrs and 11.4 yrs
      1. Total avg of whole class = n1w1+n2w2/n1+n2
      2. 14*12.2+16*11.4/30=170.8+182.4/30 = 353.2/30=11.77
      3. thus statements and b are required to find avg age of total students studying in the class.
  3. Out of a total of 10 students, the average weight of 9 students is 60 kg. If the weight of the 10th student is 9 kg more than the average of 10 students, then the weight of 10th student? – 70 kg
    1. Solution:
      1. Let the average weight of 10 students be x
      2. Total weight of 10 students = 10x
      3. Total weight of 9 students = 9 × 60 = 540 kg
      4. Weight of 10th students = (x + 9) kg
        1. According to question,
          1. ⇒ 10x – (x + 9) = 540
            1. ⇒ 9x = 549
            2. ⇒ x = 61 kg
            3. ∴ The weight of 10th  student would be 70 kg.
  4. By selling 70 pencils for rs.90 a person makes a loss of 25% how many pencils to be sold for rs. 234 at a profit of 30%
    1. 105
    1. Solution:
    1. S.P of 70 pencils = rs. 90 at loss of 25%
      1. C.P of 70 pencils = 90*100/75
      2. Therefore CP of 1 pencil = 120/70
    2. rs.234 = 130/100 * C.P @ profit 30%
      1. C.P = 180
    1. No. pencils = 180/12 * 7 = 105 pencils.
  5. The C.I on a certain sum of money for 3 years at the rate of 10% p.a is Rs.993.00 The SI on the same at the rate of 8% p.a for 4 years would be equal to
    1. Rs.960
    1. Solution:
      1. C.I = P(1+R/100)^n – P
      2. 993 = P[(1+11/10)^3 – 1]
      3. P = 993*1000/331 = rs.3000
      4. So, SI=P*n*R/100 = rs.960
      5. Reference:
  6. A 50-year-old man has a son whose age is 2/5 of his age. After how many years the ratio of the age of the man to his son will be 5/3?
    1. 25 years
      1. Solution:
    1. Given 5x = 50 years, So 2x or son’s age = 20 years
    2. Let x be the number of years to obtain the given ratio 5/3
    3. 50+x/(20+x) = 5/3
    4. = 3(50+x) = 5(20+x), Hence x = 25 years
    5. Therefore the ratio of their ages will be 5:3 in 25 years

Measurements:

  1. A solid in the form of a cylinder with hemispherical ends. The diameter of the cylinder is 1/8 of its height ( excluding the hemispherical ends) The total length of the solid including the hemispherical ends is 56 cm. What is the surface area of the solid?

    1. 2464 cm^2
      Let the radius of the cylinder and the hemisphere be r cm.
    1. Diameter of the cylinder = (2r) cm
      1. Height of the cylinder = (8 × 2r) = 16rcm
      2. Total length of the solid = 16r+r+r = 18r = 56 cm, so r = 28/9 cm
      3. The surface area of the solid hemisphere = CSA of 2 circular region + Surface area of cylinder
      4. = 2πrh + 2×2πr^2
      5. d = 1/3 h and h=3 d so, d=14 cm.
      6. 2 pir^2+2pi*r^2+pi*d*h = 4*pi*(d/2)^2+pi*d*h = pi*d^2+pi*d*h [Since 2r =d, r=d/2]
      7. pi*d+3*pi*d^2 = 4*pi*d^2 = 4*pi*14^2 =2464 cm^2
  2. The length of the side of the square (5 cm * 5 cm) is increased by 1% By what % the area of the square increases?
    1. 2%,
    2. Solution:
      1. Area of the square =5*5 = 25cm^2
      2. When the length is increased by 1%
      3. Then length of the square = 5+5*1/100 = 5.05 cm
      4. Area of new square = (5.05 * 5.05) = 25.5025 cm^2
      5. Increase in area = 25.5025-25 = 0.5025
      6. % increase in area = 0.5025/25*100 =2%
  3. A circle has the same area as that of a square of diagonal of length 11 cm. The diameter of a circle – 
    1. approx 8.7 cm
      1. Reference:–
    1. According to question,
      1. pi*r^2 = a^2
    2. By pythogoras theorem,
      1. 11^2 = a^2+a^2
      2. a= 11/root 2
    3. Area of the square = a^2 = (11/root 2)^2 = 121/2
    4. Area of the circle = pi*d^2 = (121*2)
    5. From given ,
      1. 22/7*r^2 = 60.5
      2. r^2 = 60.5/3.14 = 19.26
      3. r=4.38 cm, So diameter (2r) = 8.76cm
  4. The 2 cities P and Q are 360 km apart from each other. A car goes from P to Q with a speed of 40 km/hr and returns to P with a speed of 60 km/hr. The average speed of the car? –
    1. 48km/hr
    2. Solution:
      1. Average speed = Total Distance / Total time
      2. (2×40×60/40+60)kmph (4800/100)kmph=48 kmph
      3. Method -2:
        1. Time from P to Q = 360/40 = 9 km/hr
        2. Time from Q to P = 360/60 = 6 km/hr
        3. Total time = 9+6 = 15km/hr
        4. Avg. Speed = 720(360+360) /15 = 48km/hr.
  5. 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 it would take for him to ride both ways?
    1. Walking time be W and Riding time be R
    2. W+R = 40 mins
    3. W+W = 60 mins
    4. W = 30 mins
      1. So, R = 40-30 = 10 min
      2. Time taken for riding both ways = 10*2 = 20mins
NTA NET paper 1 Maths questions solved