CBSE Class 10 Maths Basic 2023 Question Paper 430/4/1 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/4/1 (2023),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
Assertion (A) : A tangent to a circle is perpendicular to the radius through the point of contact. Reason (R) : The lengths of tangents drawn from an external point to a circle are equal.
(A)Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.
Show answer & solution
Answer: (B) Both A and R are true but R does not give the correct explanation of A.
A is a standard theorem, so A is true.
R is also a standard theorem, so R is true.
R is about equal tangent lengths and does not explain why the tangent is perpendicular to the radius.
Assertion (A) : If one root of the quadratic equation 4x2−10x+(k−4)=0 is reciprocal of the other, then value of k is 8. Reason (R) : Roots of the quadratic equation x2−x+1=0 are real.
(A)Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.
Show answer & solution
Answer: (C) A is true but R is false.
If the roots are α and α1, their product is 1, so 4k−4=1, giving k=8. A is true.
For x2−x+1=0, D=1−4=−3<0, so the roots are not real. R is false.
A lending library has a fixed charge for first three days and an additional charge for each day thereafter. Rittik paid ₹ 27 for a book kept for 7 days and Manmohan paid ₹ 21 for a book kept for 5 days. Find the fixed charges and the charge for each extra day.
Show answer & solution
Answer: Fixed charge ₹ 15; charge for each extra day ₹ 3.
Let the fixed charge be ₹ x and the charge per extra day be ₹ y.
Two concentric circles with centre O are of radii 3 cm and 5 cm. Find the length of chord AB of the larger circle which touches the smaller circle at P.
Show answer & solution
Answer: AB = 8 cm
OP is a radius of the smaller circle and AB touches it at P, so OP⊥AB.
The perpendicular from the centre bisects the chord, so AP = PB.
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30∘ than when it was 60∘. Find the height of the tower.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60∘ and the angle of depression of its foot is 45∘. Determine the height of the tower.
Show answer & solution
Answer:7(1+3) m (about 19.12 m)
Let AB be the building (AB = 7 m) and CD the tower, with C its foot. Let the horizontal through A meet CD at E, so CE = 7 m.
Angle of depression of C is 45∘: tan45∘=BC7⇒BC=7 m, so AE = 7 m.
Angle of elevation of D is 60∘: tan60∘=AEDE⇒DE=73 m.
The following table gives the monthly consumption of electricity of 100 families : Monthly Consumption (in units): 130-140, 140-150, 150-160, 160-170, 170-180, 180-190, 190-200 Number of families: 5, 9, 17, 28, 24, 10, 7 Find the median of the above data.
The boilers are used in thermal power plants to store water and then used to produce steam. One such boiler consists of a cylindrical part in middle and two hemispherical parts at its both ends. Length of the cylindrical part is 7m and radius of cylindrical part is 27 m. Find the total surface area and the volume of the boiler. Also, find the ratio of the volume of cylindrical part to the volume of one hemispherical part.
Show answer & solution
Answer: TSA =308 m2; volume =62695≈449.17 m3; ratio =3:1
r=27 m, h=7 m.
TSA = CSA of cylinder + 2 × CSA of hemisphere =2πrh+4πr2.
=2×722×27×7+4×722×449=154+154=308 m2.
Volume of cylinder =πr2h=722×449×7=2539 m3.
Volume of two hemispheres =34πr3=34×722×8343=3539 m3.
Use of mobile screen for long hours makes your eye sight weak and give you headaches. Children who are addicted to play “PUBG” can get easily stressed out. To raise social awareness about ill effects of playing PUBG, a school decided to start ‘BAN PUBG’ campaign, in which students are asked to prepare campaign board in the shape of a rectangle. One such campaign board made by class X student of the school is shown in the figure. Based on the above information, answer the following questions : (i) Find the coordinates of the point of intersection of diagonals AC and BD. (1) (ii) Find the length of the diagonal AC. (1) (iii) (a) Find the area of the campaign Board ABCD. (2) OR (b) Find the ratio of the length of side AB to the length of the diagonal AC. (2)
Show answer & solution
Answer: (i) (4,3) (ii) 213 units (iii) (a) 24 square units OR (b) 3:13
(i) Diagonals of a rectangle bisect each other; mid-point of AC =(21+7,21+5)=(4,3).
(ii) AC=(7−1)2+(5−1)2=36+16=52=213 units.
(iii) (a) AB =7−1=6, AD =5−1=4; area =6×4=24 square units.
Khushi wants to organize her birthday party. Being health conscious, she decided to serve only fruits in her birthday party. She bought 36 apples and 60 bananas and decided to distribute fruits equally among all. Based on the above information, answer the following questions : (i) How many guests Khushi can invite at the most ? (1) (ii) How many apples and bananas will each guest get ? (1) (iii) (a) If Khushi decides to add 42 mangoes, how many guests Khushi can invite at the most ? (2) OR (b) If the cost of 1 dozen of bananas is ₹ 60, the cost of 1 apple is ₹ 15 and cost of 1 mango is ₹ 20, find the total amount spent on 60 bananas, 36 apples and 42 mangoes. (2)
Show answer & solution
Answer: (i) 12 (ii) 3 apples and 5 bananas (iii) (a) 6 OR (b) ₹ 1680
Observe the figures given below carefully and answer the questions : (i) Name the figure(s) wherein two figures are similar. (1) (ii) Name the figure(s) wherein the figures are congruent. (1) (iii) (a) Prove that congruent triangles are also similar but not the converse. (2) OR (b) What more is least needed for two similar triangles to be congruent ? (2)
Show answer & solution
Answer: (i) Figure A (and Figure C, since congruent figures are also similar) (ii) Figure C (iii) (a) Proved. OR (b) At least one pair of corresponding sides must be equal.
(i) In Figure A, corresponding angles are equal (85∘, 70∘, 105∘, 100∘) and sides are in the same ratio: 4.82.4=42=52.5=2.41.2=21, so A(i) and A(ii) are similar. In Figure B the sides are not proportional (33.5=33). Figure C triangles are congruent, hence also similar.
(ii) In Figure C, AB = PQ = 2 cm, BC = QR = 3 cm, AC = PR = 4.5 cm, so △ABC≅△PQR (SSS).
(iii) (a) If △ABC≅△PQR, then corresponding angles are equal and PQAB=QRBC=RPCA=1, so the triangles are similar. Converse: an equilateral triangle of side 1 cm and one of side 2 cm are similar (equal angles) but not congruent, since their sides are unequal.
(iii) (b) Similar triangles already have equal angles and proportional sides; if one pair of corresponding sides is equal, the ratio is 1 and all sides are equal, so they are congruent.