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Study Material for Class 10 Mathematics Mathematics Case Studies
Class 10 Mathematics students should refer to the following Pdf for Mathematics Case Studies in Class 10. These notes and test paper with questions and answers for Class 10 Mathematics will be very useful for exams and help you to score good marks
Class 10 Mathematics Mathematics Case Studies
CBSE Class 10 Mathematics Case Study 1
Case-1. Mahesh works as a manager in a hotel. He has to arrange seats in hall for a function. A hall has a certain number of chairs. Guests want to sit in different groups like in pairs, triplets, quadruplets, fives and sixes etc. When Mahesh arranges chairs in such pattern like in 2’s, 3’s, 4’s 5’s and 6’s then 1, 2, 3, 4 and 5 chairs are left respectively. But when he arranges in 11’s, no chair will be left.
(i) In the hall, how many chairs are available?
(a) 407
(b) 143
(c) 539
(d) 209
Ans: By dividing all the options by 2, 3, 4, 5, 6 and 11, we will get that 539 is the only option which leaves remainder 1, 2, 3, 4, 5, 0 respectively.
Thus (c) is correct option.
(ii) If one chair is removed, which arrangements are possible now?
(a) 2
(b) 3
(c) 4
(d) 5
Ans:-After removing 1 chair, we are left with 538 chairs. On arranging chairs in pair of 3’s, 4’s, 5’s, 6’s, 11’s ; 1, 2, 3 ,4, 10 chairs are left. So, only pair of 2 chairs is possible now.
Thus (a) is correct option.
(iii) If one chair is added to the total number of chairs, how many chairs will be left when arranged in 11’s.
(a) 1
(b) 2
(c) 3
(d) 4
Ans:-539 chairs are already arranged in pair of 11’s .On adding 1 extra chair, that 1 chair will be left only.
Thus (a) is correct option.
(iv) How many chairs will be left in original arrangement if same number of chairs will be arranged in 7’s?
(a) 0
(b) 1
(c) 2
(d) 3
Ans:-539 is divisible by 7 and remainder is zero, so arranging chairs in pair of 7’s, no chair will be left.
Thus (a) is correct option.
(v) How many chairs will be left in original arrangement if same number of chairs will be arranged in 9’s?
(a) 8
(b) 1
(c) 6
(d) 3
Ans:-539 is divisible by 9 and remainder is 8, so arranging chairs in pair of 9’s, 8 chair will be left.
Thus (a) is correct option.
CBSE Class 10 Mathematics Case Study 2
2. Indian Army is the third biggest military contingent in the World next to USA and China. However, there are many firsts that make Indian army stand out in the world, making us all Indians very proud. Knowing them, will help you celebrate Republic day with greater vigour and gratitude. On 71th republic day Parade in Delhi Captian RS Meel is planing for parade of following two group:
(a) First group of Army contingent of 624 members behind an army band of 32 members.
(b) Second group of CRPF troops with 468 soldiers behind the 228 members of bikers.
These two groups are to march in the same number of columns. This sequence of soldiers is followed by different states Jhanki which are showing the culture of the respective states.
(i) What is the maximum number of columns in which the army troop can march?
(a) 8
(b) 16
(c) 4
(d) 32
Ans:-We will find the HCF (624, 32) = 16
Thus (b) is correct option.
(ii) What is the maximum number of columns in which the CRPF troop can march?
(a) 4
(b) 8
(c) 12
(d) 16
Ans:- We will find the HCF (228, 468) = 12.
Thus (c) is correct option.
(iii) What is the maximum number of columns in which total army troop and CRPF troop together can march past?
(a) 2
(b) 4
(c) 6
(d) 8
Ans:- HCF(624, 32, 228, 468) = 4
Alternatively we can find,
HCF (16, 12) = 4
Thus (b) is correct option.
(iv) What should be subtracted with the numbers of CRPF soldiers and the number of bikers so that their maximum number of column is equal to the maximum number of column of army troop?
(a) 4 Soldiers and 4 Bikers
(b) 4 Soldiers and 2 Bikers
(c) 2 Soldiers and 4 Bikers
(d) 2 Soldiers and 2 Bikers
Ans:-Maximum number of column of army troop is 16. But 228 and 468 are not divisible by 16. If we subtract 4 from 228 and 468, both(224 and 464) are divisible by 16.
Thus (a) is correct option.
(iv) What should be added with the numbers of CRPF soldiers and the number of bikers so that their maximum number of column is equal to the maximum number of column of army troop?
(a) 4 Soldiers and 4 Bikers
(b) 12 Soldiers and 12 Bikers
(c) 6 Soldiers and 6 Bikers
(d) 12 Soldiers and 6 Bikers
Ans:-Maximum number of column of army troop is 16. But 228 and 468 are not divisible by 16. If we add 12 from 228 and 468, both(240 and 480) are divisible by 16.
Thus (b) is correct option.
CBSE Class 10 Mathematics Case Study 3
3. Shalvi wants to organize her birthday party. She was happy on her birthday. She is very health conscious, thus she decided to serve fruits only.
She has 36 apples and 60 bananas at home and decided to serve them. She want to distribute fruits among guests. She does not want to discriminate among guests so she decided to distribute equally among all.
(i) How many maximum guests Shalvi can invite?
(a) 12
(b) 120
(c) 6
(d) 180
(a) 3 apple 5 banana
(c) 2 apple 4 banana
Ans:-Out of 15 apples, each guest will get (36 ÷ 12) = 3 apples and out of 60 bananas, each guest will get (60 ÷ 12) = 5 bananas.
Thus (a) is correct option.
(iii) Shalvi decide to add 42 mangoes also. In this case how many maximum guests Shalvi can invite ?
(a) 12
(c) 6
Ans:- In this case we need to calculate HCF (36, 42, 60) =6.
Thus fruits will be equally distributed among 6 guests.
Thus (c) is correct option.
(iv) How many total fruits will each guest get?
(a) 6 apple 5 banana and 6 mangoes
(b) 6 apple 10 banana and 7 mangoes
(c) 3 apple 5 banana and 7 mangoes
(d) 3 apple 10 banana and 6 mangoes
Ans:-Out of 36 apples, each guest will get (36 ÷ 6) =6 apples and out of 42 mangoes, each guest will get (42 ÷ 6) = 7 mangoes, out of 60 bananas, each guest will get (60 ÷ 6) = 10 bananas.
Thus each guest will get 6 + 7 + 12 = 25 fruits.
Thus (b) is correct option.
(v) If Shalvi decide to add 3 more mangoes and instead 6 apple, in this case how many maximum guests Shalvi can invite ?
(a) 12
(c) 15
Ans:-Now Shalvi has 30 apples, 60 bananas, and 45 mangoes. HCF (30, 45, 60) = 15. Thus Shalvi can invite 15 guest.
Thus (c) is correct option.
4. Amar, Akbar and Anthony are playing a game. Amar climbs 5 stairs and gets down 2 stairs in one turn. Akbar goes up by 7 stairs and comes down by 2 stairs every time. Anthony goes 10 stairs up and 3 stairs down each time. Doing this they have to reach to the nearest point of 100th stairs and they will stop once they find it impossible to go forward. (They have less number of stairs than required forward stairs).
(i) Who reaches the nearest point?
(a) Amar
(b) Akbar
(c) Anthony
(d) All together reach to the nearest point.
Ans:-Amar will reach up to 93 steps then he will go for 5 steps up and 2 steps down hence covering 96 steps. Since 100 th step is final, so he will not cover more steps. Akbar will reach up to 95 steps, since 100 th step is final, so he will not cover more steps. Anthony will reach up to 91 steps, since 100 th step is final, so she will not cover more steps. Thus akbar reaches the nearest point.
Thus (b) is correct option.
(ii) How many times can they meet in between on same step?
(a) 3
(b) 4
(c) 5
(d) No, they cannot meet in between on same step.
Ans:- LCM (3, 5, 7) =105 step.
Since, total steps are 100 steps, they cannot meet in between on same step.
Thus (d) is correct option.
(iii) Who takes least number of steps to reach near hundred?
(a) Amar
(b) Akbar
(c) Anthony
(d) All of them take equal number of steps.
Ans:- Amar will take 32 steps, Akbar will take 19 steps and Anthony will take 13 steps to reach to 96 steps, 95 steps and 91 steps respectively.
Thus (c) is correct option.
(iv) What is the first stair where any two out of three will meet together?
(a) Amar and Akbar will meet for the first time after 15 steps.
(b) Akbar and Anthony will meet for the first time after 35 steps.
(b) Amar and Anthony will meet for the first time after 21 steps.
(d) Amar and Akbar will meet for the first time after 21 steps.
Ans:- Since LCM(3, 5)=15 ; LCM(5, 7)=35 ; LCM(3, 7)=21. Since, 15 is the smallest so Amar and Akbar will meet for the first time after 15 steps.
Thus (a) is correct option.
(v) What is the second stair where any two out of three will meet together?
(a) Amar and Akbar will meet after 21 steps.
(b) Akbar and Anthony will meet after 35 steps.
(b) Amar and Anthony will after 21 steps.
(b) Amar and Anthony will after 35 steps.
Ans As already calculated in (iii), LCM(3, 7) =21
Thus (b) is correct option.
CBSE Class 10 Mathematics Case Study 5
5. Ashish supplies bread and jams to a hospital and a school. Bread and jam are supplied in equal number of pieces. Bread comes in a bunch of 8 pieces and Jam comes in a pack of 6 pieces. On a particular day, Ashish has supplied x packets of bread and y packets of jam to the school. On the same day, Ashish has supplied 3x packets of bread along with sufficient packets of jam to hospital. It is known that the number of students in the school are between 500 and 550.
(i) How many students are there in school?
(a) 508
(b) 504
(c) 512
(d) 548
Ans Firstly we will find LCM (8, 6) =24. Now we will find a multiple of 24 in between 500 and 550 i.e., 504 or 528. Thus there 504 students in school.
Thus (b) is correct option.
(ii) How many packets of bread are supplied in the school?
(a) 63 packets
(b) 86 packets
(c) 65 packets
(d) 84 packets
Ans For equal distribution of bread among each student, we need 504 pieces of bread. Hence, we need (504/8=63) i.e. 63 packets of bread.
Thus (a) is correct option.
(iii) How many packets of jams are supplied in the school?
(a) 63 packets
(b) 86 packets
(c) 65 packets
(d) 84 packets
Ans For equal distribution of jam pieces among each student, we need 504 pieces of jam. Hence, we need (504/6=84) i.e. 84 packets of jam.
Thus (d) is correct option.
(iv) How many packets of bread are supplied in the hospital?
(a) 189 packets
(b) 252 packets
(c) 165 packets
(d) 288 packets
Ans For hospital, we need 3x packets of bread i.e. 3 # 63 = 189 packets of bread.
Thus (a) is correct option.
(v) How many packets of jams are supplied in the hospital?
(a) 248 packets
(b) 252 packets
(c) 165 packets
(d) 288 packets
Ans Since, number of bread pieces are 189 x 8 = 1512 , so we need same number of jam pieces. Hence 1512/6 = 252 packets of jam are distributed in the hospital.
Thus (b) is correct option.
CBSE Class 10 Mathematics Case Study 6
6. For the box to satisfy certain requirements, its length must be three meter greater than the width, and its height must be two meter less than the width.
(i) If width is taken as x , which of the following polynomial represent volume of box ?
(c) x3 - 6x2 - 6x
Ans V(x) = x(x + 3) (x − 2) = x(x2 + x − 6) = x3 + x2 − 6x
Thus (d) is correct option.
(ii) Which of the following polynomial represent the area of paper sheet used to make box ?
(a) x2 - 5x - 6
(c) x3- 6x2 - 6x
Ans S(x) = 2^LW+WH + HLh
= 2[x(x + 3) + (x + 3) (x − 2) + x(x − 2)]
= 2[x2 + 3x + x2 + x − 6 + x2 − 2x]
= 2(3x2 + 2x − 6) = 6x2 + 4x − 12
Thus (b) is correct option.
(iii) If it must have a volume of 18 unit, what must be its length ?
(a) 6 unit
(c) 4 unit
Ans We have
x3+ x2− 6x − 18 = 0
x3− 3x2+ 4x2− 12x + 6x − 18 = 0
x2 (x−3)+ 4x(x − 3)+ 6(x − 3) = 0
(x − 3)^x2+ 4x + 6h = 0
Thus width is 3 unit.
Length = x + 3 = 6 m
Thus (a) is correct option.
(iv) At a volume of 18 cubic unit, what must be its height ?
(a) 1 unit
(c) 2 unit
Ans Height = x − 2 = 3 − 2 = 1m
Thus (a) is correct option.
(v) If box is made of a paper sheet which cost is 100 rs per square unit, what is the cost of paper?
(a) Rs 5400
(c) Rs 2700
Ans S(x) = 6x2 + 4x − 12 = 6 # 3 # 3 + 4 # 3 − 12 = 54
S = 2(LW+WH + HLh
= 2(6x3 + 3x1 + 1x6)
= 2x(18 + 3 + 6) = 2x27 = 54 m2
C = 100#54 = 5400 `
Thus (a) is correct option.
(i) How many field goals did he make ?
(a) 10 Goals
(c) 15 Goals
Ans Let x be the free throw and y be the fixed goal.
As per question
x + y = 30
y = x + 10
Solving x = 10, y = 20
Thus he made 20 fixed goal.
Thus (b) is correct option.
(ii) How many free throws did he make?
(a) 10 Goals
(c) 15 Goals
Ans Free throw x = 10
Thus (a) is correct option.
(iii) What was the total number of points scored?
(a) 50
(c) 60
Ans Point scored = 10 + 2#20 = 50
Thus (a) is correct option.
(iv) If Wilt Chamberlain played 5 games during this season, what was the average number of points per game?
(a) 5
(c) 10
Ans Average point 5
50 = 10
Thus (c) is correct option.
(v) If Wilt Chamberlain played 10 games during this season, what was the average number of points per game?
(c) 4
Ans Average point 10
50 = 5
Thus (d) is correct option.
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CBSE Class 10 Mathematics Mathematics Case Studies Study Material
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